{"id":62877,"date":"2023-09-12T15:18:00","date_gmt":"2023-09-12T09:48:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=62877"},"modified":"2023-11-13T11:35:27","modified_gmt":"2023-11-13T06:05:27","slug":"rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 13 &#8211; Linear Equations In Two Variables (Updated for 2023-24)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124489\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-13-Linear-Equations-In-Two-Variables.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 13\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-13-Linear-Equations-In-Two-Variables.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-13-Linear-Equations-In-Two-Variables-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p style=\"text-align: left;\"><span style=\"font-weight: 400;\"><strong>RD Sharma Solutions Class 9 Maths Chapter 13 &#8211; Linear Equations In Two Variables:<\/strong> Are you wondering about the methods which will be effective for your performance in Class 9 Annual Exam? You don\u2019t have to worry much about that because we are going to present you <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a>&nbsp;Chapter 13.<\/span><\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d4c3fd98b08\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#download-rd-sharma-solutions-class-9-maths-chapter-13-pdf\" title=\"Download RD Sharma Solutions Class 9 Maths Chapter 13 PDF\">Download RD Sharma Solutions Class 9 Maths Chapter 13 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#exercise-wise-rd-sharma-solutions-class-9-maths-chapter-13\" title=\"Exercise-Wise RD Sharma Solutions Class 9 Maths Chapter 13\">Exercise-Wise RD Sharma Solutions Class 9 Maths Chapter 13<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#access-answers-of-rd-sharma-solutions-class-11-maths-chapter-13\" title=\"Access answers of RD Sharma Solutions Class 11 Maths Chapter 13\">Access answers of RD Sharma Solutions Class 11 Maths Chapter 13<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#rd-sharma-class-9-solutions-chapter-13-linear-equations-in-two-variables-ex-131\" title=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1\">RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#rd-sharma-solutions-class-9-chapter-13-linear-equations-in-two-variables-ex-132\" title=\"RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables Ex 13.2\">RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables Ex 13.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#linear-equations-in-two-variables-class-9-rd-sharma-solutions-ex-133\" title=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions Ex 13.3\">Linear Equations in Two Variables Class 9 RD Sharma Solutions Ex 13.3<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#rd-sharma-class-9-maths-book-questions-chapter-13-linear-equations-in-two-variables-ex-134\" title=\"RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables Ex 13.4\">RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables Ex 13.4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#linear-equations-in-two-variables-class-9-rd-sharma-solutions-vsaqs\" title=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions VSAQS\">Linear Equations in Two Variables Class 9 RD Sharma Solutions VSAQS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#rd-sharma-class-9-solution-chapter-13-linear-equations-in-two-variables-mcqs\" title=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables MCQS\">RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables MCQS<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#detailed-exercise-wise-explanation-with-important-topics\" title=\"Detailed Exercise-wise Explanation with Important Topics\">Detailed Exercise-wise Explanation with Important Topics<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#rd-sharma-solutions-class-9-chapter-13a\" title=\"RD Sharma Solutions Class 9 Chapter 13A\">RD Sharma Solutions Class 9 Chapter 13A<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#rd-sharma-solutions-class-9-chapter-13b\" title=\"RD Sharma Solutions Class 9 Chapter 13B\">RD Sharma Solutions Class 9 Chapter 13B<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#rd-sharma-solutions-class-9-chapter-13c\" title=\"RD Sharma Solutions Class 9 Chapter 13C\">RD Sharma Solutions Class 9 Chapter 13C<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#rd-sharma-solutions-class-9-chapter-13d\" title=\"RD Sharma Solutions Class 9 Chapter 13D\">RD Sharma Solutions Class 9 Chapter 13D<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#important-concepts-rd-sharma-class-9-maths\" title=\"Important concepts RD Sharma Class 9 Maths\">Important concepts RD Sharma Class 9 Maths<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-13\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 13\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 13<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-17\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-13\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 13?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 13?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-13\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 13?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 13?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-19\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-13-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 13\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 13\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-9-maths-chapter-13-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths Chapter 13 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/RD-Sharma-Class-9-Maths-Chapter-13.pdf\">RD Sharma Solutions Class 9 Maths Chapter 13<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/RD-Sharma-Class-9-Maths-Chapter-13.pdf\", \"#example1\");<\/script><\/p>\n<h2 style=\"text-align: left;\"><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-9-maths-chapter-13\"><\/span><strong>Exercise-Wise RD Sharma Solutions Class 9 Maths Chapter 13<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-13-class-9-maths-exercise-13-1-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 13 Exercise 13.1<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-13-class-9-maths-exercise-13-2-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 13 Exercise 13.2<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-13-class-9-maths-exercise-13-3-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 13 Exercise 13.3<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-13-class-9-maths-exercise-13-4-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 13 Exercise 13.4<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-11-maths-chapter-13\"><\/span><strong>Access answers of RD Sharma Solutions Class 11 Maths Chapter 13<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solutions-chapter-13-linear-equations-in-two-variables-ex-131\"><\/span>RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>Three angles of a quadrilateral are respectively equal to 110\u00b0, 50\u00b0 and 40\u00b0. Find its fourth angle.<br>Solution:<br>Sum of four angles of a quadrilateral = 360\u00b0<br>Three angles are = 110\u00b0, 50\u00b0 and 40\u00b0<br>\u2234 Fourth angle = 360\u00b0 \u2013 (110\u00b0 + 50\u00b0 + 40\u00b0)<br>= 360\u00b0 \u2013 200\u00b0 = 160\u00b0<\/p>\n<p>Question 2.<br>In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measures of each angle of the quadrilateral.<br>Solution:<br>Sum of angles of a quadrilateral ABCD = 360\u00b0<br>Ratio in angles = 1 : 2 : 4 : 5<br>Let first angle = x<br>Second angle = 2x<br>Third angle = 4x<br>and fourth angle = 5x<br>\u2234 x + 2x + 4x + 5x = 360\u00b0<br>\u21d2 12x = 360\u00b0 \u21d2 x =&nbsp;<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"msubsup\"><span id=\"MathJax-Span-5\" class=\"texatom\"><span id=\"MathJax-Span-6\" class=\"mrow\"><span id=\"MathJax-Span-7\" class=\"mn\">360<\/span><\/span><\/span><span id=\"MathJax-Span-8\" class=\"texatom\"><span id=\"MathJax-Span-9\" class=\"mrow\"><span id=\"MathJax-Span-10\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-11\" class=\"mn\">12<\/span><\/span><\/span><\/span><\/span>&nbsp; = 30\u00b0<br>\u2234 First angle = 30\u00b0<br>Second angle = 30\u00b0 x 2 = 60\u00b0<br>Third angle = 30\u00b0 x 4 = 120\u00b0<br>Fourth angle = 30\u00b0 x 5 = 150\u00b0<\/p>\n<p>Question 3.<br>The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. [NCERT]<br>Solution:<br>Sum of four angles of a quadrilateral = 360\u00b0<br>Ratio in the angles = 3 : 5 : 9 : 13<br>Let first angle = 3x<br>Then second angle = 5x<br>Third angle = 9x<br>and fourth angle = 13x<br>\u2234 3x + 5x + 9x+ 13x = 360\u00b0<br>\u21d2 30x = 360\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-12\" class=\"math\"><span id=\"MathJax-Span-13\" class=\"mrow\"><span id=\"MathJax-Span-14\" class=\"mfrac\"><span id=\"MathJax-Span-15\" class=\"msubsup\"><span id=\"MathJax-Span-16\" class=\"texatom\"><span id=\"MathJax-Span-17\" class=\"mrow\"><span id=\"MathJax-Span-18\" class=\"mn\">360<\/span><\/span><\/span><span id=\"MathJax-Span-19\" class=\"texatom\"><span id=\"MathJax-Span-20\" class=\"mrow\"><span id=\"MathJax-Span-21\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-22\" class=\"mn\">30<\/span><\/span><\/span><\/span><\/span>&nbsp;= 12\u00b0<br>\u2234 First angle = 3x = 3 x 12\u00b0 = 36\u00b0<br>Second angle = 5x = 5 x 12\u00b0 = 60\u00b0<br>Third angle = 9x = 9 x 12\u00b0 = 108\u00b0<br>Fourth angle = 13 x 12\u00b0 = 156\u00b0<\/p>\n<p>Question 4.<br>In a quadrilateral ABCD, CO and DO are the bisectors of \u2220C and \u2220D respectively.<br>Prove that \u2220COD =&nbsp;<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-23\" class=\"math\"><span id=\"MathJax-Span-24\" class=\"mrow\"><span id=\"MathJax-Span-25\" class=\"mfrac\"><span id=\"MathJax-Span-26\" class=\"mn\">1<\/span><span id=\"MathJax-Span-27\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;(\u2220A + \u2220B).<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1970\/45594027692_f8f1e55f14_o.png\" alt=\"RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"354\" height=\"395\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1939\/45594027172_ded69e05ed_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"358\" height=\"225\"><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-13-linear-equations-in-two-variables-ex-132\"><\/span>RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables Ex 13.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>Two opposite angles of a parallelogram are (3x- 2)\u00b0 and (50 \u2013 x)\u00b0. Find the measure of each angle of the parallelogram.<br>Solution:<br>\u2235 Opposite angles of a parallelogram are equal<br>\u2234 3x \u2013 2 = 50 \u2013 x<br>\u21d2 3x + x \u2013 50 + 2<br>\u21d2 4x = 52<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-28\" class=\"math\"><span id=\"MathJax-Span-29\" class=\"mrow\"><span id=\"MathJax-Span-30\" class=\"mfrac\"><span id=\"MathJax-Span-31\" class=\"mn\">52<\/span><span id=\"MathJax-Span-32\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;= 13<br>\u2234 \u2220A = 3x \u2013 2 = 3 x 13 \u2013 2 = 39\u00b0 \u2013 2 = 37\u00b0<br>\u2220C = 50\u00b0 -x = 50\u00b0 \u2013 13 = 37\u00b0<br>But\u2220A + B = 180\u00b0<br>\u2234 37\u00b0 + \u2220B = 180\u00b0<br>\u21d2 \u2220B = 180\u00b0 \u2013 37\u00b0 = 143\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1921\/45594027042_b246019af2_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables\" width=\"199\" height=\"151\"><br>and \u2220D = \u2220B (Opposite angles of a ||gm)<br>\u2234 \u2220D = 143\u00b0<br>Hence angles and 37\u00b0, 143\u00b0, 37\u00b0, 143\u00b0<\/p>\n<p>Question 2.<br>If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.<br>Solution:<br>Let in ||gm ABCD,<br>\u2220A =x<br>Then \u2220B =&nbsp;<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-33\" class=\"math\"><span id=\"MathJax-Span-34\" class=\"mrow\"><span id=\"MathJax-Span-35\" class=\"mfrac\"><span id=\"MathJax-Span-36\" class=\"mn\">2<\/span><span id=\"MathJax-Span-37\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;x<br><img src=\"https:\/\/farm2.staticflickr.com\/1960\/45644578311_397e9e6daf_o.png\" alt=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions\" width=\"188\" height=\"160\"><br>But, \u2220A + \u2220B = 180\u00b0 (Sum of two adjacent angles of a ||gm)<br>\u21d2 x +&nbsp;<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-38\" class=\"math\"><span id=\"MathJax-Span-39\" class=\"mrow\"><span id=\"MathJax-Span-40\" class=\"mfrac\"><span id=\"MathJax-Span-41\" class=\"mn\">2<\/span><span id=\"MathJax-Span-42\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>x = 180\u00b0<br>\u21d2&nbsp;<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-43\" class=\"math\"><span id=\"MathJax-Span-44\" class=\"mrow\"><span id=\"MathJax-Span-45\" class=\"mfrac\"><span id=\"MathJax-Span-46\" class=\"mn\">5<\/span><span id=\"MathJax-Span-47\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>x = 180\u00b0<br>\u21d2 x = 180\u00b0 x&nbsp;<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-48\" class=\"math\"><span id=\"MathJax-Span-49\" class=\"mrow\"><span id=\"MathJax-Span-50\" class=\"mfrac\"><span id=\"MathJax-Span-51\" class=\"mn\">3<\/span><span id=\"MathJax-Span-52\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp;= 108\u00b0<br>\u2234 \u2220A = 108\u00b0<br>and \u2220B = 108\u00b0 x&nbsp;<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-53\" class=\"math\"><span id=\"MathJax-Span-54\" class=\"mrow\"><span id=\"MathJax-Span-55\" class=\"mfrac\"><span id=\"MathJax-Span-56\" class=\"mn\">2<\/span><span id=\"MathJax-Span-57\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;= 72\u00b0<br>But, \u2220A = \u2220C and \u2220B = \u2220D (Opposite angles of a ||gm)<br>\u2234 \u2220C = 108\u00b0, \u2220D = 72\u00b0<br>Hence angles are 108\u00b0, 72\u00b0, 108\u00b0, 72\u00b0<\/p>\n<p>Question 3.<br>Find the measure of all the angles of a parallelogram, if one angle is 24\u00b0 less than twice the smallest angle.<br>Solution:<br>Let smallest angle of a ||gm = x<br>Then second angle = 2x \u2013 24\u00b0<br>But these are consecutive angles<br>\u2234 x + (2x- 24\u00b0) = 180\u00b0<br>\u21d2 x + 2x \u2013 24\u00b0 = 180\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1942\/45594026762_2ccc2a9b34_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables\" width=\"212\" height=\"161\"><br>\u21d2 3x = 180\u00b0 + 24\u00b0 = 204\u00b0<br>\u21d2 x =<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-58\" class=\"math\"><span id=\"MathJax-Span-59\" class=\"mrow\"><span id=\"MathJax-Span-60\" class=\"mfrac\"><span id=\"MathJax-Span-61\" class=\"msubsup\"><span id=\"MathJax-Span-62\" class=\"texatom\"><span id=\"MathJax-Span-63\" class=\"mrow\"><span id=\"MathJax-Span-64\" class=\"mn\">204<\/span><\/span><\/span><span id=\"MathJax-Span-65\" class=\"texatom\"><span id=\"MathJax-Span-66\" class=\"mrow\"><span id=\"MathJax-Span-67\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-68\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp; = 68\u00b0<br>\u2234 Smallest angle = 68\u00b0<br>and second angle = 2x 68\u00b0 \u2013 24\u00b0<br>= 136\u00b0-24\u00b0 = 112\u00b0<br>\u2235 The opposite angles of a ||gm are equal Other two angles will be 68\u00b0 and 112\u00b0<br>\u2234 Hence angles are 68\u00b0, 112\u00b0, 68\u00b0, 112\u00b0<\/p>\n<p>Question 4.<br>The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?<br>Solution:<br>In a ||gm ABC,<br><img src=\"https:\/\/farm2.staticflickr.com\/1964\/45644578151_af1920a358_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"197\" height=\"157\"><br>Perimeter = 22cm<br>and longest side = 6.5 cm<br>Let shorter side = x<br>\u2234 2x (6.5 + x) = 22<br>\u21d2 13 + 2x = 22<br>\u21d2 2x = 22 \u2013 13 = 9<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-69\" class=\"math\"><span id=\"MathJax-Span-70\" class=\"mrow\"><span id=\"MathJax-Span-71\" class=\"mfrac\"><span id=\"MathJax-Span-72\" class=\"mn\">9<\/span><span id=\"MathJax-Span-73\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 4.5<br>\u2234 shorter side = 4.5cm<\/p>\n<p>Question 5.<br>In a parallelogram ABCD, \u2220D = 135\u00b0, determine the measure of \u2220A and \u2220B.<br>Solution:<br>In ||gm ABCD,<br>\u2220D = 135\u00b0<br>But, \u2220A + \u2220D = 180\u00b0 (Sum of consecutive angles)<br>\u21d2\u2220A+ 135\u00b0 = 180\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1972\/45644577911_780642a02f_o.png\" alt=\"Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions\" width=\"205\" height=\"142\"><br>\u21d2 \u2220A = 180\u00b0 \u2013 135\u00b0 = 45\u00b0<br>\u2235 \u2220B = \u2220D (Opposite angles of a ||gm)<br>\u2234 \u2220B = 135\u00b0<\/p>\n<p>Question 6.<br>ABCD is a parallelogram in which \u2220A = 70\u00b0. Compute \u2220B, \u2220C and \u2220D.<br>Solution:<br>In ||gm ABCD,<br>\u2220A = 70\u00b0<br>But \u2220A + \u2220B = 180\u00b0 (Sum of consecutive angles)<br>\u21d2 70\u00b0 + \u2220B = 180\u00b0<br>\u21d2 \u2220B = 180\u00b0 \u2013 70\u00b0 = 110\u00b0<br>But \u2220C = \u2220A and \u2220D = \u2220B (Opposite angles of a ||gm)<br>\u2220C = 70\u00b0 and \u2220D = 110\u00b0<br>Hence \u2220B = 110\u00b0, \u2220C = 70\u00b0 and \u2220D = 110\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1932\/45644577701_5a877a13fc_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables\" width=\"184\" height=\"142\"><\/p>\n<p>Question 7.<br>In the figure, ABCD is a parallelogram in which \u2220DAB = 75\u00b0 and \u2220DBC = 60\u00b0. Compute \u2220CDB and \u2220ADB.<br><img src=\"https:\/\/farm2.staticflickr.com\/1968\/45644577531_f1a894cbbc_o.png\" alt=\"RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables\" width=\"186\" height=\"146\"><br>Solution:<br>In ||gm ABCD,<br>\u2220A + \u2220B = 180\u00b0<br>(Sum of consecutive angles) But, \u2220A = 75\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1933\/45644577361_2329b0cab0_o.png\" alt=\"Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions\" width=\"190\" height=\"147\"><br>\u2234 \u2220B = 180\u00b0 \u2013 \u2220A = 180\u00b0 \u2013 75\u00b0 = 105\u00b0<br>\u2234 DBA = 105\u00b0 -60\u00b0 = 45\u00b0<br>But \u2220CDB = \u2220DBA (alternate angles)<br>= 45\u00b0<br>and \u2220ADB = \u2220DBC = 60\u00b0<\/p>\n<p>Question 8.<br>Which of the following statements are true (T) and which are false (F)?<br>(i) In a parallelogram, the diagonals are equal.<br>(ii) In a parallelogram, the diagonals bisect each other.<br>(iii) In a parallelogram, the diagonals intersect each other at right angles.<br>(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.<br>(v) If all the angles of a quadrilateral are equal, it is a parallelogram.<br>(vi) If three sides of a quadrilateral are equal, it is a parallelogram.<br>(vii) If three angles of a quadrilateral are equal, it is a parallelogram.<br>(viii)If all the sides of a quadrilateral are equal it is a parallelogram.<br>Solution:<br>(i) False, Diagonals of a parallelogram are not equal.<br>(ii) True.<br>(iii) False, Diagonals bisect each other at right angles is a rhombus or a square only.<br>(iv) False, In a quadrilateral, if opposite sides are equal and parallel, then it is a ||gm.<br>(v) False, If all angles are equal, then it is a square or a rectangle.<br>(vi) False, If opposite sides are equal and parallel then it is a ||gm<br>(vii) False, If opposite angles are equal, then it is a parallelogram.<br>(viii)False, If all the sides are equal then it is a square or a rhombus but not parallelogram.<\/p>\n<p>Question 9.<br>In the figure, ABCD is a parallelogram in which \u2220A = 60\u00b0. If the bisectors of \u2220A and \u2220B meet at P, prove that AD = DP, PC= BC and DC = 2AD.<br><img src=\"https:\/\/farm2.staticflickr.com\/1905\/45644577161_02c54a7431_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables\" width=\"168\" height=\"150\"><br>Solution:<br>Given : In ||gm ABCD,<br>\u2220A = 60\u00b0<br>Bisector of \u2220A and \u2220B meet at P.<br>To prove :<br>(i) AD = DP<br>(ii) PC = BC<br>(iii) DC = 2AD<br>Construction : Join PD and PC<br>Proof : In ||gm ABCD,<br>\u2220A = 60\u00b0<br>But \u2220A + \u2220B = 180\u00b0 (Sum of excutive angles)<br>\u21d2 60\u00b0 + \u2220B = 180\u00b0<br>\u2234 \u2220B = 1809 \u2013 60\u00b0 = 120\u00b0<br>\u2235 DC || AB<br>\u2220PAB = \u2220DPA (alternate angles)<br>\u21d2 \u2220PAD = \u2220DPA (\u2235 \u2220PAB = \u2220PAD)<br>\u2234 AB = DP<br>(PA is its angle bisector, sides opposite to equal angles)<br>(ii) Similarly, we can prove that \u2220PBC = \u2220PCB (\u2235 \u2220PAB = \u2220BCA alternate angles)<br>\u2234 PC = BC<br>(iii) DC = DP + PC<br>= AD + BC [From (i) and (ii)]<br>= AD + AB = 2AB (\u2235BC = AD opposite sides of the ||gm)<br>Hence DC = 2AD<\/p>\n<p>Question 10.<br>In the figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove thatAF = 2AB.<br>Solution:<br>Given : In ||gm ABCD,<br>E a mid point of BC<br>DE is joined and produced to meet AB produced at F<br><img src=\"https:\/\/farm2.staticflickr.com\/1933\/45644576881_85b4b01da1_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"198\" height=\"123\"><br>To prove : AF = 2AB<br>Proof : In \u2206CDE and \u2206EBF<br>\u2220DEC = \u2220BEF (vertically opposite angles)<br>CE = EB (E is mid point of BC)<br>\u2220DCE = \u2220EBF (alternate angles)<br>\u2234 \u2206CDE \u2245 \u2206EBF (SAS Axiom)<br>\u2234 DC = BF (c.p.c.t.)<br>But AB = DC (opposite sides of a ||gm)<br>\u2234 AB = BF<br>Now, AF = AB + BF = AB + AB = 2AB<br>Hence AF = 2AB<\/p>\n<h3><span class=\"ez-toc-section\" id=\"linear-equations-in-two-variables-class-9-rd-sharma-solutions-ex-133\"><\/span>Linear Equations in Two Variables Class 9 RD Sharma Solutions Ex 13.3<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In a parallelogram ABCD, determine the sum of angles ZC and ZD.<br>Solution:<br>In a ||gm ABCD,<br>\u2220C + \u2220D = 180\u00b0<br>(Sum of consecutive angles)<br><img src=\"https:\/\/farm2.staticflickr.com\/1973\/31772237468_92bb8a5799_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables\" width=\"175\" height=\"156\"><\/p>\n<p>Question 2.<br>In a parallelogram ABCD, if \u2220B = 135\u00b0, determine the measures of its other angles.<br>Solution:<br>In a ||gm ABCD, \u2220B = 135\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1908\/31772237278_aeaee77da5_o.png\" alt=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions\" width=\"205\" height=\"152\"><br>\u2234 \u2220D = \u2220B = 135\u00b0 (Opposite angles of a ||gm)<br>But \u2220A + \u2220B = 180\u00b0 (Sum of consecutive angles)<br>\u21d2 \u2220B + 135\u00b0 = 180\u00b0<br>\u2234 \u2220A = 180\u00b0 \u2013 135\u00b0 = 45\u00b0<br>But\u2220C = \u2220B = 45\u00b0 (Opposite angles of a ||gm)<br>\u2234 Angles are 45\u00b0, 135\u00b0, 45\u00b0, 135\u00b0.<\/p>\n<p>Question 3.<br>ABCD is a square, AC and BD intersect at O. State the measure of \u2220AOB.<br>Solution:<br>In a square ABCD,<br>Diagonal AC and BD intersect each other at O<br>\u2235 Diagonals of a square bisect each other at right angle<br>\u2235\u2220AOB = 90\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1937\/31772237078_af7242d371_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables\" width=\"154\" height=\"161\"><\/p>\n<p>Question 4.<br>ABCD is a rectangle with \u2220ABD = 40\u00b0. Determine \u2220DBC.<br>Solution:<br>In rectangle ABCD,<br><img src=\"https:\/\/farm2.staticflickr.com\/1902\/31772236878_62efb7e80b_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"185\" height=\"155\"><br>\u2220B = 90\u00b0, BD is its diagonal<br>But \u2220ABD = 40\u00b0<br>and \u2220ABD + \u2220DBC = 90\u00b0<br>\u21d2 40\u00b0 + \u2220DBC = 90\u00b0<br>\u21d2 \u2220DBC = 90\u00b0 \u2013 40\u00b0 = 50\u00b0<br>Hence \u2220DBC = 50\u00b0<\/p>\n<p>Question 5.<br>The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.<br>Solution:<br>Given : In ||gm ABCD, E and F are the mid points of the side AB and CD respectively<br>DE and BF are joined<br>To prove : EBFD is a ||gm<br>Construction : Join EF<br><img src=\"https:\/\/farm2.staticflickr.com\/1903\/44730426115_c07826bc97_o.png\" alt=\"Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions\" width=\"187\" height=\"152\"><br>Proof : \u2235 ABCD is a ||gm<br>\u2234 AB = CD and AB || CD<br>(Opposite sides of a ||gm are equal and parallel)<br>\u2234 EB || DF and EB = DF (\u2235 E and F are mid points of AB and CD)<br>\u2234 EBFD is a ||gm.<\/p>\n<p>Question 6.<br>P and Q are the points of trisection of the diagonal BD of the parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.<br>Solution:<br>Given : In ||gm, ABCD. P and Q are the points of trisection of the diagonal BD<br><img src=\"https:\/\/farm2.staticflickr.com\/1913\/44730425925_0bef5152ea_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables\" width=\"188\" height=\"153\"><br>To prove : (i) CQ || AP<br>AC bisects PQ<br>Proof: \u2235 Diagonals of a parallelogram bisect each other<br>\u2234 AO = OC and BO = OD<br>\u2234 P and Q are point of trisection of BD<br>\u2234 BP = PQ = QD \u2026(i)<br>\u2235 BO = OD and BP = QD \u2026(ii)<br>Subtracting, (ii) from (i) we get<br>OB \u2013 BP = OD \u2013 QD<br>\u21d2 OP = OQ<br>In quadrilateral APCQ,<br>OA = OC and OP = OQ (proved)<br>Diagonals AC and PQ bisect each other at O<br>\u2234 APCQ is a parallelogram<br>Hence AP || CQ.<\/p>\n<p>Question 7.<br>ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.<br>Solution:<br>Given : In square ABCD<br>E, F, G and H are the points on AB, BC, CD and DA respectively such that AE = BF = CG = DH<br>To prove : EFGH is a square<br>Proof : E, F, G and H are points on the sides AB, BC, CA and DA respectively such that<br>AE = BF = CG = DH = x (suppose)<br>Then BE = CF = DG = AH = y (suppose)<br>Now in \u2206AEH and \u2206BFE<br><img src=\"https:\/\/farm2.staticflickr.com\/1943\/44730425695_faf719fc75_o.png\" alt=\"RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables\" width=\"182\" height=\"162\"><br>AE = BF (given)<br>\u2220A = \u2220B (each 90\u00b0)<br>AH = BE (proved)<br>\u2234 \u2206AEH \u2245 \u2206BFE (SAS criterion)<br>\u2234 \u22201 = \u22202 and \u22203 = \u22204 (c.p.c.t.)<br>But \u22201 + \u22203 = 90\u00b0 and \u22202 + \u22204 = 90\u00b0 (\u2220A = \u2220B = 90\u00b0)<br>\u21d2 \u22201 + \u22202 + \u22203 + \u22204 = 90\u00b0 + 90\u00b0 = 180\u00b0<br>\u21d2 \u22201 + \u22204 + \u22201 + \u22204 = 180\u00b0<br>\u21d2 2(\u22201 + \u22204) = 180\u00b0<br>\u21d2 \u22201 + \u22204 =&nbsp;<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-74\" class=\"math\"><span id=\"MathJax-Span-75\" class=\"mrow\"><span id=\"MathJax-Span-76\" class=\"mfrac\"><span id=\"MathJax-Span-77\" class=\"msubsup\"><span id=\"MathJax-Span-78\" class=\"texatom\"><span id=\"MathJax-Span-79\" class=\"mrow\"><span id=\"MathJax-Span-80\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-81\" class=\"texatom\"><span id=\"MathJax-Span-82\" class=\"mrow\"><span id=\"MathJax-Span-83\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-84\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 90\u00b0<br>\u2234 \u2220HEF = 180\u00b0 \u2013 90\u00b0 = 90\u00b0<br>Similarly, we can prove that<br>\u2220F = \u2220G = \u2220H = 90\u00b0<br>Since sides of quad. EFGH is are equal and each angle is of 90\u00b0<br>\u2234 EFGH is a square.<\/p>\n<p>Question 8.<br>ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.<br>Solution:<br>Given : ABCD is a rhombus, EABF is a straight line such that<br>EA = AB = BF<br>ED and FC are joined<br>Which meet at G on producing<br><img src=\"https:\/\/farm2.staticflickr.com\/1908\/44730425425_7ed746a390_o.png\" alt=\"Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions\" width=\"257\" height=\"150\"><br>To prove: \u2220EGF = 90\u00b0<br>Proof : \u2235 Diagonals of a rhombus bisect<br>each other at right angles<br>AO = OC, BO = OD<br>\u2220AOD = \u2220COD = 90\u00b0<br>and \u2220AOB = \u2220BOC = 90\u00b0<br>In \u2206BDE,<br>A and O are the mid-points of BE and BD respectively.<br>\u2234 AO || ED<br>Similarly, OC || DG<br>In \u2206 CFA, B and O are the mid-points of AF and AC respectively<br>\u2234 OB || CF and OD || GC<br>Now in quad. DOCG<br>OC || DG and OD || CG<br>\u2234 DOCG is a parallelogram.<br>\u2234 \u2220DGC = \u2220DOC (opposite angles of ||gm)<br>\u2234 \u2220DGC = 90\u00b0 (\u2235 \u2220DOC = 90\u00b0)<br>Hence proved.<\/p>\n<p>Question 9.<br>ABCD is a parallelogram, AD is produced to E so that DE = DC = AD and EC produced meets AB produced in F. Prove that BF = BC.<br>Solution:<br>Given : In ||gm ABCD,<br>AB is produced to E so that<br>DE = DA and EC produced meets AB produced in F.<br>To prove : BF = BC<br>Proof: In \u2206ACE,<br><img src=\"https:\/\/farm2.staticflickr.com\/1961\/44730425395_c3a3c0fe19_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables\" width=\"150\" height=\"168\"><br>O and D are the mid points of sides AC and AE<br>\u2234 DO || EC and DB || FC<br>\u21d2 BD || EF<br>\u2234 AB = BF<br>But AB = DC (Opposite sides of ||gm)<br>\u2234 DC = BF<br>Now in \u2206EDC and \u2206CBF,<br>DC = BF (proved)<br>\u2220EDC = \u2220CBF<br>(\u2235\u2220EDC = \u2220DAB corresponding angles)<br>\u2220DAB = \u2220CBF (corresponding angles)<br>\u2220ECD = \u2220CFB (corresponding angles)<br>\u2234 AEDC \u2245 ACBF (ASA criterion)<br>\u2234 DE = BC (c.p.c.t.)<br>\u21d2 DC = BC<br>\u21d2 AB = BC<br>\u21d2 BF = BC (\u2235AB = BF proved)<br>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-maths-book-questions-chapter-13-linear-equations-in-two-variables-ex-134\"><\/span>RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables Ex 13.4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In a \u2206ABC, D, E and F are respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7cm, 8cm and 9cm, respectively, find the perimeter of \u2206DEF.<br>Solution:<br>In \u2206ABC, D, E and F are the mid-points of sides,<br>BC, CA, AB respectively<br>AB = 7cm, BC = 8cm and CA = 9cm<br>\u2235 D and E are the mid points of BC and CA<br>\u2234 DE || AB and DE =<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-85\" class=\"math\"><span id=\"MathJax-Span-86\" class=\"mrow\"><span id=\"MathJax-Span-87\" class=\"mfrac\"><span id=\"MathJax-Span-88\" class=\"mn\">1<\/span><span id=\"MathJax-Span-89\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AB =<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-90\" class=\"math\"><span id=\"MathJax-Span-91\" class=\"mrow\"><span id=\"MathJax-Span-92\" class=\"mfrac\"><span id=\"MathJax-Span-93\" class=\"mn\">1<\/span><span id=\"MathJax-Span-94\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;x 7 = 3.5cm<br>Similarly,<br><img src=\"https:\/\/farm2.staticflickr.com\/1931\/44730425365_eddfaf5c27_o.png\" alt=\"RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"294\" height=\"346\"><\/p>\n<p>Question 2.<br>In a triangle \u2220ABC, \u2220A = 50\u00b0, \u2220B = 60\u00b0 and \u2220C = 70\u00b0. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.<br>Solution:<br>In \u2206ABC,<br>\u2220A = 50\u00b0, \u2220B = 60\u00b0 and \u2220C = 70\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1969\/44730425025_c87fcbbbe9_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"252\" height=\"170\"><br>D, E and F are the mid points of sides BC, CA and AB respectively<br>DE, EF and ED are joined<br>\u2235 D, E and F are the mid points of sides BC, CA and AB respectively<br>\u2234 EF || BC<br>DE || AB and FD || AC<br>\u2234 BDEF and CDEF are parallelogram<br>\u2234 \u2220B = \u2220E = 60\u00b0 and \u2220C = \u2220F = 70\u00b0<br>Then \u2220A = \u2220D = 50\u00b0<br>Hence \u2220D = 50\u00b0, \u2220E = 60\u00b0 and \u2220F = 70\u00b0<\/p>\n<p>Question 3.<br>In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29cm and AB = 30cm, find the perimeter of the quadrilateral ARPQ.<br>Solution:<br>P, Q, R are the mid points of sides BC, CA and AB respectively<br>AC = 21 cm, BC = 29 cm and AB = 30\u00b0<br>\u2235 P, Q, R and the mid points of sides BC, CA and AB respectively.<br>\u2234 PQ || AB and PQ =&nbsp;<span id=\"MathJax-Element-15-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-95\" class=\"math\"><span id=\"MathJax-Span-96\" class=\"mrow\"><span id=\"MathJax-Span-97\" class=\"mfrac\"><span id=\"MathJax-Span-98\" class=\"mn\">1<\/span><span id=\"MathJax-Span-99\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AB<br><img src=\"https:\/\/farm2.staticflickr.com\/1958\/44730424885_9a0ed779c9_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"265\" height=\"232\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1915\/44730424795_40e17dd689_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables\" width=\"302\" height=\"327\"><\/p>\n<p>Question 4.<br>In a \u2206ABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.<br>Solution:<br>Given : In \u2206ABC, AD is median and AD is produced to X such that DX = AD<br>To prove : ABXC is a parallelogram<br>Construction : Join BX and CX<br>Proof : In \u2206ABD and \u2206CDX<br>AD = DX (Given)<br>BD = DC (D is mid points)<br>\u2220ADB = \u2220CDX (Vertically opposite angles)<br>\u2234 \u2206ABD \u2245 \u2206CDX (SAS criterian)<br>\u2234 AB = CX (c.p.c.t.)<br>and \u2220ABD = \u2220DCX<br>But these are alternate angles<br><img src=\"https:\/\/farm2.staticflickr.com\/1903\/44730424655_81e6523691_o.png\" alt=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions\" width=\"213\" height=\"254\"><br>\u2234 AB || CX and AB = CX<br>\u2234 ABXC is a parallelogram.<\/p>\n<p>Question 5.<br>In a \u2206ABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.<br>Solution:<br>Given : In \u2206ABC, E and F are the mid-points of AC and AB respectively.<br>EF are joined.<br>AP \u22a5 BC is drawn which intersects EF at Q and meets BC at P.<br>To prove: AQ = QP<br>proof : In \u2206ABC<br><img src=\"https:\/\/farm2.staticflickr.com\/1914\/44730424515_924d4895f1_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables\" width=\"251\" height=\"175\"><br>E and F are the mid points of AC and AB<br>\u2234 EF || BC and EF =&nbsp;<span id=\"MathJax-Element-16-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-100\" class=\"math\"><span id=\"MathJax-Span-101\" class=\"mrow\"><span id=\"MathJax-Span-102\" class=\"mfrac\"><span id=\"MathJax-Span-103\" class=\"mn\">1<\/span><span id=\"MathJax-Span-104\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>BC<br>\u2234 \u2220F = \u2220B<br>In \u2206ABP,<br>F is mid point of AB and Q is the mid point of FE or FQ || BC<br>\u2234 Q is mid point of AP,<br>\u2234 AQ = QP<\/p>\n<p>Question 6.<br>In a \u2206ABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML NL.<br>Solution:<br>In \u2206ABC,<br>BM and CN are perpendicular on a line drawn from A. L is the mid point of BC. ML and NL are joined.<br><img src=\"https:\/\/farm2.staticflickr.com\/1945\/44730424445_004c386a59_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"196\" height=\"261\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1977\/44730424405_55f9c96401_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 - 6a\" width=\"355\" height=\"549\"><\/p>\n<p>Question 7.<br>In the figure triangle ABC is right-angled at B. Given that AB = 9cm. AC = 15cm and D, E are the mid points of the sides AB and AC respectively, calculate.<br>(i) The length of BC<br>(ii) The area of \u2206ADC<br>Solution:<br>In \u2206ABC, \u2220B = 90\u00b0<br>AC =15 cm, AB = 9cm<br>D and E are the mid points of sides AB and AC respectively and D, E are joined.<br><img src=\"https:\/\/farm2.staticflickr.com\/1945\/44730424225_79ed4183e6_o.png\" alt=\"Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions\" width=\"353\" height=\"728\"><\/p>\n<p>Question 8.<br>In the figure, M, N and P are the mid points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5cm and MP = 2.5cm, calculate BC, AB and AC.<br><img src=\"https:\/\/farm2.staticflickr.com\/1968\/44730423915_6f58122277_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables\" width=\"179\" height=\"175\"><br>Solution:<br>In \u2206ABC,<br>M, N and P are the mid points of side, AB, AC and BC respectively.<br><img src=\"https:\/\/farm2.staticflickr.com\/1914\/44730423835_0e7ae9e22a_o.png\" alt=\"RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables\" width=\"360\" height=\"606\"><\/p>\n<p>Question 9.<br>In the figure, AB = AC and CP || BA and AP is the bisector of exterior \u2220CAD of \u2206ABC. Prove that (i) \u2220PAC = \u2220BCA (ii) ABCP is a parallelogram.<br>Solution:<br>Given : In ABC, AB = AC<br><img src=\"https:\/\/farm2.staticflickr.com\/1905\/44730423615_b6c844570b_o.png\" alt=\"Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions\" width=\"213\" height=\"197\"><br>nd CP || BA, AP is the bisector of exterior \u2220CAD of \u2206ABC<br>To prove :<br>(i) \u2220PAC = \u2220BCA<br>(ii) ABCP is a ||gm<br>Proof : (i) In \u2206ABC,<br>\u2235 AB =AC<br>\u2234 \u2220C = \u2220B (Angles opposite to equal sides) and ext.<br>\u2220CAD = \u2220B + \u2220C<br>= \u2220C + \u2220C = 2\u2220C \u2026.(i)<br>\u2235 AP is the bisector of \u2220CAD<br>\u2234 2\u2220PAC = \u2220CAD \u2026(ii)<br>From (i) and (ii)<br>\u2220C = 2\u2220PAC<br>\u2220C = \u2220CAD or \u2220BCA = \u2220PAC<br>Hence \u2220PAC = \u2220BCA<br>(ii) But there are alternate angles,<br>\u2234 AD || BC<br>But BA || CP<br>\u2234 ABCP is a ||gm.<\/p>\n<p>Question 10.<br>ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.<br>Solution:<br>Given : In fne figure, ABCD is a kite in which AB = AD and BC = CD.<br>P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.<br>To prove : PQRS is a rectangle.<br>Construction : Join AC and BD.<br><img src=\"https:\/\/farm2.staticflickr.com\/1978\/44730423445_73165e3959_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables\" width=\"206\" height=\"264\"><br>Proof: In \u2206ABD,<br>P and S are mid points of AB and AD<br>\u2234 PS || BD and PS =&nbsp;<span id=\"MathJax-Element-17-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-105\" class=\"math\"><span id=\"MathJax-Span-106\" class=\"mrow\"><span id=\"MathJax-Span-107\" class=\"mfrac\"><span id=\"MathJax-Span-108\" class=\"mn\">1<\/span><span id=\"MathJax-Span-109\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;BD \u2026(i)<br>Similarly in \u2206BCD,<br>Q and R the mid points of BC and CD<br>\u2234 QR || BD and<br>QR =&nbsp;<span id=\"MathJax-Element-18-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-110\" class=\"math\"><span id=\"MathJax-Span-111\" class=\"mrow\"><span id=\"MathJax-Span-112\" class=\"mfrac\"><span id=\"MathJax-Span-113\" class=\"mn\">1<\/span><span id=\"MathJax-Span-114\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;BD \u2026(ii)<br>\u2234 Similarly, we can prove that PQ || SR and PQ = SR \u2026(iii)<br>From (i) and (ii) and (iii)<br>PQRS is a parallelogram,<br>\u2235 AC and BD intersect each other at right angles.<br>\u2234 PQRS is a rectangle.<\/p>\n<p>Question 11.<br>Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid-points of the sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.<br>Solution:<br>In \u2206ABC, AB = AC<br>D, E and F are the mid points of the sides BC, CA and AB respectively,<br>AD and EF are joined intersecting at O<br>To prove : AD and EF bisect each other at right angles.<br>Construction : Join DE and DF.<br><img src=\"https:\/\/farm2.staticflickr.com\/1916\/43826453580_6d6dd3b19d_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"168\" height=\"174\"><br>Proof : \u2235 D, E and F are the mid-points of<br>the sides BC, CA and AB respectively<br>\u2234 AFDE is a ||gm<br>\u2234 AF = DE and AE = DF<br>But AF = AE<br>(\u2235 E and F are mid-points of equal sides AB and AC)<br>\u2234 AF = DF = DE = AE<br>\u2234AFDE is a rhombus<br>\u2235 The diagonals of a rhombus bisect each other at right angle.<br>\u2234 AO = OD and EO = OF<br>Hence, AD and EF bisect each other at right angles.<\/p>\n<p>Question 12.<br>Show that the line segments joining the mid points of the opposite sides of a quadrilateral bisect each other.<br>Solution:<br>Given : In quad. ABCD,<br>P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively.<br>PR and QS to intersect each other at O<br>To prove : PO = OR and QO = OS<br>Construction: Join PQ, QR, RS and SP and also join AC.<br>Proof: In \u2206ABC<br>P and Q are mid-points of AB and BC<br>\u2234 PQ || AC and PQ =&nbsp;<span id=\"MathJax-Element-19-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-115\" class=\"math\"><span id=\"MathJax-Span-116\" class=\"mrow\"><span id=\"MathJax-Span-117\" class=\"mfrac\"><span id=\"MathJax-Span-118\" class=\"mn\">1<\/span><span id=\"MathJax-Span-119\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AC \u2026(i)<br>Similarly is \u2206ADC,<br>S and R are the mid-points of AD and CD<br>\u2234 SR || AC and SR =&nbsp;<span id=\"MathJax-Element-20-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-120\" class=\"math\"><span id=\"MathJax-Span-121\" class=\"mrow\"><span id=\"MathJax-Span-122\" class=\"mfrac\"><span id=\"MathJax-Span-123\" class=\"mn\">1<\/span><span id=\"MathJax-Span-124\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AC ..(ii)<br><img src=\"https:\/\/farm2.staticflickr.com\/1939\/44730423215_301d3db1da_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"175\" height=\"179\"><br>from (i) and (ii)<br>PQ = SQ and PQ || SR<br>PQRS is a ||gm (\u2235 opposite sides are equal area parallel)<br>But the diagonals of a ||gm bisect each other.<br>\u2234 PR and QS bisect each other.<\/p>\n<p>Question 13.<br>Fill in the blanks to make the following statements correct :<br>(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is \u2026<br>(ii) The triangle formed by joining the mid-points of the sides of a right triangle is \u2026<br>(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is \u2026<br>Solution:<br>(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is an isosceles triangle.<br><img src=\"https:\/\/farm2.staticflickr.com\/1969\/43826453400_16ff84798c_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"183\" height=\"167\"><br>(ii) The triangle formed by joining the mid-points of the sides of a right triangle is right triangle.<br><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44730423065_764ffab47f_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 13 Linear Equations in Two Variables\" width=\"132\" height=\"174\"><br>(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is a parallelogram.<br><img src=\"https:\/\/farm2.staticflickr.com\/1951\/43826453220_357dc200cc_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"182\" height=\"182\"><\/p>\n<p>Question 14.<br>ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of \u2206PQR is double the perimeter of \u2206ABC.<br>Solution:<br>Given : In \u2206ABC,<br>Through A, B and C, lines are drawn parallel to BC, CA and AB respectively meeting at P, Q and R.<br><img src=\"https:\/\/farm2.staticflickr.com\/1960\/44730422895_d17f7caced_o.png\" alt=\"RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"229\" height=\"183\"><br>To prove : Perimeter of \u2206PQR = 2 x perimeter of \u2206ABC<br>Proof : \u2235 PQ || BC and QR || AB<br>\u2234 ABCQ is a ||gm<br>\u2234 BC = AQ<br>Similarly, BCAP is a ||gm<br>\u2234 BC = AP \u2026(i)<br>\u2234 AQ = AP = BL<br>\u21d2 PQ = 2BC<br>Similarly, we can prove that<br>QR = 2AB and PR = 2AC<br>Now perimeter of \u2206PQR.<br>= PQ + QR + PR = 2AB + 2BC + 2AC<br>= 2(AB + BC + AC)<br>= 2 perimeter of \u2206ABC.<br>Hence proved<\/p>\n<p>Question 15.<br>In the figure, BE \u22a5 AC. AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that PQR = 90\u00b0.<br>Solution:<br>Given: In \u2206ABC, BE \u22a5 AC<br>AD is any line from A to BC meeting BC in D and intersecting BE in H. P, Q and R are respectively mid points of AH, AB and BC. PQ and QR are joined B.<br><img src=\"https:\/\/farm2.staticflickr.com\/1937\/43826453050_1dfb7a3a70_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"220\" height=\"174\"><br>To prove : \u2220PQR = 90\u00b0<br>Proof: In \u2206ABC,<br>Q and R the mid points of AB and BC 1<br>\u2234 QR || AC and QR =&nbsp;<span id=\"MathJax-Element-21-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-125\" class=\"math\"><span id=\"MathJax-Span-126\" class=\"mrow\"><span id=\"MathJax-Span-127\" class=\"mfrac\"><span id=\"MathJax-Span-128\" class=\"mn\">1<\/span><span id=\"MathJax-Span-129\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AC<br>Similarly, in \u2206ABH,<br>Q and P are the mid points of AB and AH<br>\u2234 QP || BH or QP || BE<br>But AC \u22a5 BE<br>\u2234 QP \u22a5 QR<br>\u2234 \u2220PQR = 90\u00b0<\/p>\n<p>Question 16.<br>ABC is a triangle. D is a point on AB such that AD =&nbsp;<span id=\"MathJax-Element-22-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-130\" class=\"math\"><span id=\"MathJax-Span-131\" class=\"mrow\"><span id=\"MathJax-Span-132\" class=\"mfrac\"><span id=\"MathJax-Span-133\" class=\"mn\">1<\/span><span id=\"MathJax-Span-134\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;AB and E is a point on AC such that AE =&nbsp;<span id=\"MathJax-Element-23-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-135\" class=\"math\"><span id=\"MathJax-Span-136\" class=\"mrow\"><span id=\"MathJax-Span-137\" class=\"mfrac\"><span id=\"MathJax-Span-138\" class=\"mn\">1<\/span><span id=\"MathJax-Span-139\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;AC. Prove that DE =&nbsp;<span id=\"MathJax-Element-24-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-140\" class=\"math\"><span id=\"MathJax-Span-141\" class=\"mrow\"><span id=\"MathJax-Span-142\" class=\"mfrac\"><span id=\"MathJax-Span-143\" class=\"mn\">1<\/span><span id=\"MathJax-Span-144\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;BC.<br>Solution:<br>Given : In \u2206ABC,<br>D is a point on AB such that<br>AD =&nbsp;<span id=\"MathJax-Element-25-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-145\" class=\"math\"><span id=\"MathJax-Span-146\" class=\"mrow\"><span id=\"MathJax-Span-147\" class=\"mfrac\"><span id=\"MathJax-Span-148\" class=\"mn\">1<\/span><span id=\"MathJax-Span-149\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;AB and E is a point on AC such 1<br>that AE =&nbsp;<span id=\"MathJax-Element-26-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-150\" class=\"math\"><span id=\"MathJax-Span-151\" class=\"mrow\"><span id=\"MathJax-Span-152\" class=\"mfrac\"><span id=\"MathJax-Span-153\" class=\"mn\">1<\/span><span id=\"MathJax-Span-154\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;AC<br>DE is joined.<br><img src=\"https:\/\/farm2.staticflickr.com\/1975\/44919533644_b8f39aaa22_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"247\" height=\"165\"><br>To prove : DE =&nbsp;<span id=\"MathJax-Element-27-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-155\" class=\"math\"><span id=\"MathJax-Span-156\" class=\"mrow\"><span id=\"MathJax-Span-157\" class=\"mfrac\"><span id=\"MathJax-Span-158\" class=\"mn\">1<\/span><span id=\"MathJax-Span-159\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;BC<br>Construction : Take P and Q the mid points of AB and AC and join them<br>Proof: In \u2206ABC,<br>\u2235 P and Q are the mid-points of AB and AC<br><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44919533464_72bb952b2f_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables\" width=\"354\" height=\"276\"><\/p>\n<p>Question 17.<br>In the figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ =&nbsp;<span id=\"MathJax-Element-28-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-160\" class=\"math\"><span id=\"MathJax-Span-161\" class=\"mrow\"><span id=\"MathJax-Span-162\" class=\"mfrac\"><span id=\"MathJax-Span-163\" class=\"mn\">1<\/span><span id=\"MathJax-Span-164\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.<br><img src=\"https:\/\/farm2.staticflickr.com\/1915\/43826452710_728564a439_o.png\" alt=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions\" width=\"199\" height=\"161\"><br>Solution:<br>Given : In ||gm ABCD,<br>P is the mid-point of DC and Q is a point on AC such that CQ =&nbsp;<span id=\"MathJax-Element-29-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-165\" class=\"math\"><span id=\"MathJax-Span-166\" class=\"mrow\"><span id=\"MathJax-Span-167\" class=\"mfrac\"><span id=\"MathJax-Span-168\" class=\"mn\">1<\/span><span id=\"MathJax-Span-169\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;AC. PQ is produced meets BC at R.<br><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44919533114_7f0ca04e8c_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables\" width=\"201\" height=\"154\"><br>To prove : R is mid point of BC<br>Construction : Join BD<br>Proof : \u2235 In ||gm ABCD,<br>\u2235 Diagonal AC and BD bisect each other at O<br>\u2234 AO = OC =&nbsp;<span id=\"MathJax-Element-30-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-170\" class=\"math\"><span id=\"MathJax-Span-171\" class=\"mrow\"><span id=\"MathJax-Span-172\" class=\"mfrac\"><span id=\"MathJax-Span-173\" class=\"mn\">1<\/span><span id=\"MathJax-Span-174\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AC \u2026(i)<br>In \u2206OCD,<br>P and Q the mid-points of CD and CO<br>\u2234 PQ || OD and PQ =&nbsp;<span id=\"MathJax-Element-31-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-175\" class=\"math\"><span id=\"MathJax-Span-176\" class=\"mrow\"><span id=\"MathJax-Span-177\" class=\"mfrac\"><span id=\"MathJax-Span-178\" class=\"mn\">1<\/span><span id=\"MathJax-Span-179\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;OD<br>In \u2206BCD,<br>P is mid-poiht of DC and PQ || OD (Proved above)<br>Or PR || BD<br>\u2234 R is mid-point BC.<\/p>\n<p>Question 18.<br>In the figure, ABCD and PQRC are rectangles and Q is the mid-point of AC.<br>Prove that (i) DP = PC (ii) PR =&nbsp;<span id=\"MathJax-Element-32-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-180\" class=\"math\"><span id=\"MathJax-Span-181\" class=\"mrow\"><span id=\"MathJax-Span-182\" class=\"mfrac\"><span id=\"MathJax-Span-183\" class=\"mn\">1<\/span><span id=\"MathJax-Span-184\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AC.<br><img src=\"https:\/\/farm2.staticflickr.com\/1959\/43826452500_68b2592948_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"219\" height=\"168\"><br>Solution:<br>Given : ABCD are PQRC are rectangles and Q is the mid-point of AC.<br>To prove : (i) DP = PC (ii) PR =&nbsp;<span id=\"MathJax-Element-33-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-185\" class=\"math\"><span id=\"MathJax-Span-186\" class=\"mrow\"><span id=\"MathJax-Span-187\" class=\"mfrac\"><span id=\"MathJax-Span-188\" class=\"mn\">1<\/span><span id=\"MathJax-Span-189\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AC<br>Construction : Join diagonal AC which passes through Q and join PR.<br><img src=\"https:\/\/farm2.staticflickr.com\/1943\/44919532684_7f582e57bb_o.png\" alt=\"Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions\" width=\"209\" height=\"160\"><br>Proof : (i) In \u2206ACD,<br>Q is mid-point of AC and QP || AD (Sides of rectangles)<br>\u2234 P is mid-point of CD<br>\u2234 DP = PC<br>(ii) \u2235PR and QC are the diagonals of rectangle PQRC<br>\u2234 PR = QC<br>But Q is the mid-point of AC<br>\u2234 QC =&nbsp;<span id=\"MathJax-Element-34-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-190\" class=\"math\"><span id=\"MathJax-Span-191\" class=\"mrow\"><span id=\"MathJax-Span-192\" class=\"mfrac\"><span id=\"MathJax-Span-193\" class=\"mn\">1<\/span><span id=\"MathJax-Span-194\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AC<br>Hence PR =&nbsp;<span id=\"MathJax-Element-35-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-195\" class=\"math\"><span id=\"MathJax-Span-196\" class=\"mrow\"><span id=\"MathJax-Span-197\" class=\"mfrac\"><span id=\"MathJax-Span-198\" class=\"mn\">1<\/span><span id=\"MathJax-Span-199\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AC<\/p>\n<p>Question 19.<br>ABCD is a parallelogram, E and F are the mid points AB and CD respectively. GFI is any line intersecting AD, EF and BC at Q P and H respectively. Prove that GP = PH.<br>Solution:<br>Given : In ||gm ABCD,<br>E and F are mid-points of AB and CD<br>GH is any line intersecting AD, EF and BC at GP and H respectively<br><img src=\"https:\/\/farm2.staticflickr.com\/1967\/44730422275_995bffb9a9_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables\" width=\"229\" height=\"163\"><br>To prove : GP = PH<br>Proof: \u2235 E and F are the mid-points of AB and CD<br><img src=\"https:\/\/farm2.staticflickr.com\/1924\/30703633437_7ebbbf3d29_o.png\" alt=\"RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables\" width=\"353\" height=\"508\"><\/p>\n<p>Question 20.<br>BM and CN are perpendiculars to a line passing, through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.<br>Solution:<br>In \u2206ABC,<br>BM and CN are perpendicular on a line drawn from A.<br>L is the mid point of BC.<br>ML and NL are joined.<br><img src=\"https:\/\/farm2.staticflickr.com\/1928\/44730422125_915a379f7a_o.png\" alt=\"Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions\" width=\"344\" height=\"737\"><\/p>\n<h3><span class=\"ez-toc-section\" id=\"linear-equations-in-two-variables-class-9-rd-sharma-solutions-vsaqs\"><\/span>Linear Equations in Two Variables Class 9 RD Sharma Solutions VSAQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In a parallelogram ABCD, write the sum of angles A and B.<br>Solution:<br>In ||gm ABCD,<br>\u2220A + \u2220B = 180\u00b0<br>(Sum of consecutive angles of a ||gm)<br><img src=\"https:\/\/farm2.staticflickr.com\/1933\/30703633217_0844cd2190_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"181\" height=\"148\"><\/p>\n<p>Question 2.<br>In a parallelogram ABCD, if \u2220D = 115\u00b0, then write the measure of \u2220A.<br>Solution:<br>In ||gm ABCD,<br>\u2220D = 115\u00b0<br>But \u2220A + \u2220D = 180\u00b0<br>(Sum of consecutive angles of a ||gm)<br><img src=\"https:\/\/farm2.staticflickr.com\/1921\/30703633147_a8fd44cac7_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"163\" height=\"135\"><br>\u21d2 \u2220A + 115\u00b0= 180\u00b0 \u2220A = 180\u00b0- 115\u00b0<br>\u2234 \u2220A = 65\u00b0<\/p>\n<p>Question 3.<br>PQRS is a square such that PR and SQ intersect at O. State the measure of \u2220POQ.<br>Solution:<br>In a square PQRS,<br>Diagonals PR and QS intersects each other at O.<br><img src=\"https:\/\/farm2.staticflickr.com\/1970\/44730421825_bf1d810ac6_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 13 Linear Equations in Two Variables\" width=\"166\" height=\"163\"><br>\u2235 The diagonals of a square bisect each other at right angles.<br>\u2234 \u2220POQ = 90\u00b0<\/p>\n<p>Question 4.<br>If PQRS is a square then write the measure of \u2220SRP.<br>Solution:<br>In square PQRS,<br><img src=\"https:\/\/farm2.staticflickr.com\/1965\/30703633037_b5ea145027_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"165\" height=\"157\"><br>Join PR,<br>\u2235Diagonals of a square bisect are opposite angles<br>\u2234\u2220SRP =&nbsp;<span id=\"MathJax-Element-36-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-200\" class=\"math\"><span id=\"MathJax-Span-201\" class=\"mrow\"><span id=\"MathJax-Span-202\" class=\"mfrac\"><span id=\"MathJax-Span-203\" class=\"mn\">1<\/span><span id=\"MathJax-Span-204\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>x \u2220SRQ<br>=&nbsp;<span id=\"MathJax-Element-37-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-205\" class=\"math\"><span id=\"MathJax-Span-206\" class=\"mrow\"><span id=\"MathJax-Span-207\" class=\"mfrac\"><span id=\"MathJax-Span-208\" class=\"mn\">1<\/span><span id=\"MathJax-Span-209\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;x 90\u00b0 = 45\u00b0<\/p>\n<p>Question 5.<br>If ABCD is a rhombus with \u2220ABC = 56\u00b0, find the measure of \u2220ACD.<br>Solution:<br>In rhombus ABCD,<br>Diagonals bisect each other at 0 at right angles.<br>\u2220ABC = 56\u00b0<br>But \u2220ABC + \u2220BCD = 180\u00b0 (Sum of consecutive angles)<br>\u21d2 56\u00b0 + \u2220BCD = 180\u00b0<br>\u21d2 \u2220BCD = 180\u00b0 \u2013 56\u00b0 = 124\u00b0<br>\u2235 Diagonals of a rhombus bisect the opposite angle<br>\u2234 \u2220ACD =&nbsp;<span id=\"MathJax-Element-38-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-210\" class=\"math\"><span id=\"MathJax-Span-211\" class=\"mrow\"><span id=\"MathJax-Span-212\" class=\"mfrac\"><span id=\"MathJax-Span-213\" class=\"mn\">1<\/span><span id=\"MathJax-Span-214\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220BCD =&nbsp;<span id=\"MathJax-Element-39-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-215\" class=\"math\"><span id=\"MathJax-Span-216\" class=\"mrow\"><span id=\"MathJax-Span-217\" class=\"mfrac\"><span id=\"MathJax-Span-218\" class=\"mn\">1<\/span><span id=\"MathJax-Span-219\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;x 124\u00b0<br>= 62\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1971\/44730421605_7c884959fa_o.png\" alt=\"RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"187\" height=\"149\"><\/p>\n<p>Question 6.<br>The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm, what is the measure of the shorter side.<br>Solution:<br>Perimeter of a ||gm ABCD = 22cm<br>\u2234 Sum of two consecutive sides =&nbsp;<span id=\"MathJax-Element-40-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-220\" class=\"math\"><span id=\"MathJax-Span-221\" class=\"mrow\"><span id=\"MathJax-Span-222\" class=\"mfrac\"><span id=\"MathJax-Span-223\" class=\"mn\">22<\/span><span id=\"MathJax-Span-224\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br>= 11cm<br>i.e. AB + BC = 11 cm<br>AB = 6.5 cm and let BC = x cm<br>\u2234 6.5 + x = 11 cm<br>x = 11 \u2013 6.5 = 4.5<br>\u2234 Shorter side = 4.5 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1978\/30703632937_4171717735_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"197\" height=\"148\"><\/p>\n<p>Question 7.<br>If the angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Then find the measure of the smallest angle.<br>Solution:<br>Ratio in the angles of a quadrilateral = 3 : 5 : 9 : 13<br>Let first angle = 3x<br>Second angle = 5x<br>Third angle = 9x<br>and fourth angle = 13x<br>\u2235 The sum of angles of a quadrilateral = 360\u00b0<br>\u2234 3x + 5x + 9x + 13x = 360\u00b0<br>\u21d2 30x = 360\u00b0 \u21d2 x =&nbsp;<span id=\"MathJax-Element-41-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-225\" class=\"math\"><span id=\"MathJax-Span-226\" class=\"mrow\"><span id=\"MathJax-Span-227\" class=\"mfrac\"><span id=\"MathJax-Span-228\" class=\"msubsup\"><span id=\"MathJax-Span-229\" class=\"texatom\"><span id=\"MathJax-Span-230\" class=\"mrow\"><span id=\"MathJax-Span-231\" class=\"mn\">360<\/span><\/span><\/span><span id=\"MathJax-Span-232\" class=\"texatom\"><span id=\"MathJax-Span-233\" class=\"mrow\"><span id=\"MathJax-Span-234\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-235\" class=\"mn\">30<\/span><\/span><\/span><\/span><\/span>&nbsp; = 12<br>\u2234 Smallest angle = 3x = 3 x 12\u00b0 = 36\u00b0<\/p>\n<p>Question 8.<br>In parallelogram ABCD if \u2220A = (3x \u2013 20\u00b0), \u2220B = (y + 15)\u00b0, \u2220C = (x + 40\u00b0), then find the value of x and y.<br>Solution:<br>In a ||gm ABCD,<br>\u2220A = (3x \u2013 20\u00b0), \u2220B = y + 15\u00b0,<br>\u2220C = x + 40\u00b0<br>Now, \u2220A = \u2220C (Opposite angles of a ||gm)<br>\u21d2 3x \u2013 20 = x + 40\u00b0<br>\u21d2 3x \u2013 x = 40\u00b0 + 20\u00b0 \u21d2 2x = 60\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-42-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-236\" class=\"math\"><span id=\"MathJax-Span-237\" class=\"mrow\"><span id=\"MathJax-Span-238\" class=\"mfrac\"><span id=\"MathJax-Span-239\" class=\"msubsup\"><span id=\"MathJax-Span-240\" class=\"texatom\"><span id=\"MathJax-Span-241\" class=\"mrow\"><span id=\"MathJax-Span-242\" class=\"mn\">60<\/span><\/span><\/span><span id=\"MathJax-Span-243\" class=\"texatom\"><span id=\"MathJax-Span-244\" class=\"mrow\"><span id=\"MathJax-Span-245\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-246\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 30\u00b0<br>and \u2220A + \u2220B = 180\u00b0 (Sum of the consecutive angles)<br>\u21d2 3x-20\u00b0 + y + 15\u00b0 = 180\u00b0<br>\u21d2 3x + y \u2013 5\u00b0 = 180\u00b0<br>\u21d2 3 x 30\u00b0 +y- 5\u00b0 = 180\u00b0<br>\u21d2 90\u00b0 \u2013 5\u00b0 + y = 180<br>y = 180\u00b0 \u2013 90\u00b0 + 5 = 95\u00b0<br>\u2234 x = 30\u00b0, y = 95\u00b0<\/p>\n<p>Question 9.<br>If measures opposite angles of a parallelogram are (60 \u2013 x)\u00b0 and (3x \u2013 4)\u00b0, then find the measures of angles of the parallelogram.<br>Solution:<br>Opposite angles of a ||gm ABCD are (60 \u2013 x)\u00b0 and (3x \u2013 4\u00b0)<br>But opposite angles of a ||gm are equal, the<br>60\u00b0 \u2013 x\u00b0 = 3x \u2013 4\u00b0 \u21d2 60\u00b0 + 4\u00b0 = 3x + x<br>\u21d2 4x = 64\u00b0 \u21d2 x =&nbsp;<span id=\"MathJax-Element-43-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-247\" class=\"math\"><span id=\"MathJax-Span-248\" class=\"mrow\"><span id=\"MathJax-Span-249\" class=\"mfrac\"><span id=\"MathJax-Span-250\" class=\"msubsup\"><span id=\"MathJax-Span-251\" class=\"texatom\"><span id=\"MathJax-Span-252\" class=\"mrow\"><span id=\"MathJax-Span-253\" class=\"mn\">64<\/span><\/span><\/span><span id=\"MathJax-Span-254\" class=\"texatom\"><span id=\"MathJax-Span-255\" class=\"mrow\"><span id=\"MathJax-Span-256\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-257\" class=\"msubsup\"><span id=\"MathJax-Span-258\" class=\"texatom\"><span id=\"MathJax-Span-259\" class=\"mrow\"><span id=\"MathJax-Span-260\" class=\"mn\">4<\/span><\/span><\/span><span id=\"MathJax-Span-261\" class=\"texatom\"><span id=\"MathJax-Span-262\" class=\"mrow\"><span id=\"MathJax-Span-263\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>&nbsp; = 16\u00b0<br>\u2234 \u2220A = 60\u00b0 \u2013 x = 60\u00b0 \u2013 16\u00b0 = 44\u00b0<br>But \u2220A + \u2220B = 180\u00b0 (sum of consecutive angle)<br>\u21d2 44\u00b0 + \u2220B = 180\u00b0<br>\u21d2 \u2220B = 180\u00b0 \u2013 44\u00b0<br>\u21d2 \u2220B = 136\u00b0<br>But \u2220A = \u2220C and \u2220B = \u2220D (Opposite angles)<br>\u2234 Angles are 44\u00b0, 136\u00b0, 44\u00b0, 136\u00b0<\/p>\n<p>Question 10.<br>In a parallelogram ABCD, the bisectors of \u2220A also bisect BC at x, find AB : AD.<br>Solution:<br>In ||gm ABCD,<br>Bisectors of \u2220A meets BC at X and BX = XC<br>Draw XY ||gm AB meeting AD at Y<br><img src=\"https:\/\/farm2.staticflickr.com\/1980\/44730421485_85c04513b0_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"296\" height=\"494\"><\/p>\n<p>Question 11.<br>In the figure, PQRS in an isosceles trapezium find x and y.<br>Solution:<br>\u2235 PQRS is an isosceles trapezium in which<br>SP = RQ and SR || PQ<br>\u2234 \u2220P + \u2220S = 180\u00b0 (Sum of co-interior angles)<br>3x + 2x = 180\u00b0 \u21d2 5x = 180\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-44-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-264\" class=\"math\"><span id=\"MathJax-Span-265\" class=\"mrow\"><span id=\"MathJax-Span-266\" class=\"mfrac\"><span id=\"MathJax-Span-267\" class=\"msubsup\"><span id=\"MathJax-Span-268\" class=\"texatom\"><span id=\"MathJax-Span-269\" class=\"mrow\"><span id=\"MathJax-Span-270\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-271\" class=\"texatom\"><span id=\"MathJax-Span-272\" class=\"mrow\"><span id=\"MathJax-Span-273\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-274\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp; = 36\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1918\/44730421295_946869323d_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables\" width=\"207\" height=\"144\"><br>But \u2220P = \u2220Qm (Base angles of isosceles trapezium)<br>y = 2x = 2 x 36\u00b0 = 12\u00b0<br>\u2234 y = 12\u00b0<br>Hence x = 36\u00b0, y = 12\u00b0<\/p>\n<p>Question 12.<br>In the figure ABCD is a trapezium. Find the values of x and y.<br><img src=\"https:\/\/farm2.staticflickr.com\/1951\/44730421225_84b6d336b1_o.png\" alt=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions\" width=\"200\" height=\"148\"><br>Solution:<br>In trapezium ABCD,<br>AB || CD<br><img src=\"https:\/\/farm2.staticflickr.com\/1908\/44919530674_78a3b91089_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables\" width=\"198\" height=\"145\"><br>\u2234 \u2220A + \u2220D = 180\u00b0 (Sum of cointerior angles)<br>x + 20\u00b0 + 2x + 10\u00b0 = 180\u00b0<br>3x + 30\u00b0 = 180\u00b0<br>\u21d2 3x= 180\u00b0 \u2013 30\u00b0<br>3x = 150\u00b0<br>x =&nbsp;<span id=\"MathJax-Element-45-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-275\" class=\"math\"><span id=\"MathJax-Span-276\" class=\"mrow\"><span id=\"MathJax-Span-277\" class=\"mfrac\"><span id=\"MathJax-Span-278\" class=\"msubsup\"><span id=\"MathJax-Span-279\" class=\"texatom\"><span id=\"MathJax-Span-280\" class=\"mrow\"><span id=\"MathJax-Span-281\" class=\"mn\">150<\/span><\/span><\/span><span id=\"MathJax-Span-282\" class=\"texatom\"><span id=\"MathJax-Span-283\" class=\"mrow\"><span id=\"MathJax-Span-284\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-285\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp; = 50\u00b0<br>Similarly, \u2220B + \u2220C = 180\u00b0<br>\u21d2 y + 92\u00b0 = 180\u00b0<br>\u21d2 y = 180\u00b0 \u2013 92\u00b0 = 88\u00b0<br>\u2234 x = 50\u00b0, y = 88\u00b0<\/p>\n<p>Question 13.<br>In the figure, ABCD and AEFG are two parallelograms. If \u2220C = 58\u00b0, find \u2220F.<br><img src=\"https:\/\/farm2.staticflickr.com\/1965\/44919530494_f18cbe56b9_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"197\" height=\"155\"><br>Solution:<br>In the figure, ABCD and AEFG are two parallelograms \u2220C = 58\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1921\/44919530414_250044bd58_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"200\" height=\"153\"><br>\u2235 DC || GF and CB || FE (Sides of ||gms)<br>\u2234 \u2220C = \u2220F<br>But \u2220C = 58\u00b0<br>\u2234 \u2220F = 58\u00b0<\/p>\n<p>Question 14.<br>Complete each of the following statements by means of one of those given in brackets against each:<br>(i) If one pair of opposite sides are equal and parallel, then the figure is \u2026\u2026\u2026 (parallelogram, rectangle, trapezium)<br>(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is \u2026\u2026.. (square, rectangle, trapezium)<br>(iii) A line drawn from the mid-point of one side of a triangle \u2026\u2026\u2026. another side intersects the third side at its mid-point, (perpendicular to, parallel to, to meet)<br>(iv) If one angle of a parallelogram is a right angle, then it is necessarily a \u2026\u2026.. (rectangle, square, rhombus)<br>(v) Consecutive angle of a parallelogram are \u2026\u2026\u2026 (supplementary, complementary)<br>(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a \u2026\u2026\u2026 (rectangle, parallelogram, rhombus)<br>(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a \u2026\u2026\u2026. (parallelogram, rhombus, rectangle)<br>(viii)If consecutive sides of a parallelogram are equal, then it is necessarily a \u2026\u2026.. (kite, rhombus, square)<br>Solution:<br>(i) If one pair of opposite sides are equal and parallel, then the figure is parallelogram.<br>(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is trapezium.<br>(iii) A line drawn from the mid-point of one side of a triangle parallel to another side intersects the third side at its mid-point,<br>(iv) If one angle of a parallelogram is a right angle, then it is necessarily a rectangle.<br>(v) Consecutive angle of a parallelogram are supplementary.<br>(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a parallelogram.<br>(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a parallelogram.<br>(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a rhombus.<\/p>\n<p>Question 15.<br>In a quadrilateral ABCD, bisectors of A and B intersect at O such that \u2220AOB = 75\u00b0, then write the value of \u2220C + \u2220D.<br>Solution:<br>In quadrilateral ABCD,<br>Bisectors of \u2220A and \u2220B meet at O and \u2220AOB = 75\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1973\/44730420845_0385ab9592_o.png\" alt=\"Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions\" width=\"173\" height=\"178\"><br>In AOB, \u2220AOB = 75\u00b0<br>\u2234 \u2220OAB + \u2220OBA = 180\u00b0 \u2013 75\u00b0 = 105\u00b0<br>But OA and OB are the bisectors of \u2220A and \u2220B.<br>\u2234 \u2220A + \u2220B = 2 x 105\u00b0 = 210\u00b0<br>But \u2220A + \u2220B + \u2220C + \u2220D = 360\u00b0 (Sum of angles of a quad.)<br>\u2234 210\u00b0 + \u2220C + \u2220D = 360\u00b0<br>\u21d2 \u2220C + \u2220D = 360\u00b0 \u2013 210\u00b0 = 150\u00b0<br>Hence \u2220C + \u2220D = 150\u00b0<\/p>\n<p>Question 16.<br>The diagonals of a rectangle ABCD meet at O. If \u2220BOC = 44\u00b0 find \u2220OAD.<br>Solution:<br>In rectangle ABCD,<br>Diagonals AC and BD intersect each other at O and \u2220BOC = 44\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1945\/44919530094_943a540e5a_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables\" width=\"188\" height=\"158\"><br>But \u2220AOD = \u2220BOC (Vertically opposite angles)<br>\u2234 \u2220AOD = 44\u00b0<br>In \u2206AOD,<br>\u2220AOD + \u2220OAD + \u2220ODA = 180\u00b0 (Sum of angles of a triangle)<br>\u21d2 44\u00b0 + \u2220OAD + \u2220OAD = 180\u00b0 [\u2235 OA = OD, \u2220OAD = \u2220ODA]<br>\u21d2 2\u2220OAD = 180\u00b0 \u2013 44\u00b0 = 136\u00b0<br>\u2234 \u2220OAD =&nbsp;<span id=\"MathJax-Element-46-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-286\" class=\"math\"><span id=\"MathJax-Span-287\" class=\"mrow\"><span id=\"MathJax-Span-288\" class=\"mfrac\"><span id=\"MathJax-Span-289\" class=\"msubsup\"><span id=\"MathJax-Span-290\" class=\"texatom\"><span id=\"MathJax-Span-291\" class=\"mrow\"><span id=\"MathJax-Span-292\" class=\"mn\">136<\/span><\/span><\/span><span id=\"MathJax-Span-293\" class=\"texatom\"><span id=\"MathJax-Span-294\" class=\"mrow\"><span id=\"MathJax-Span-295\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-296\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 68\u00b0<\/p>\n<p>Question 17.<br>If ABCD is a rectangle with \u2220BAC = 32\u00b0, find the measure if \u2220DBC.<br>Solution:<br>In rectangle ABCD,<br><img src=\"https:\/\/farm2.staticflickr.com\/1963\/44730420505_8938ec2ebe_o.png\" alt=\"RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables\" width=\"193\" height=\"160\"><br>Diagonals bisect each other at O<br>\u2220BAC = 32\u00b0<br>\u2235 OA = OB<br>\u2234 \u2220OBA Or \u2220DBA = \u2220BAC = 32\u00b0<br>But \u2220ABC = 90\u00b0 (Angle of a rectangles)<br>\u2234 \u2220DBC = \u2220ABC \u2013 \u2220DBA<br>= 90\u00b0 \u2013 32\u00b0 = 58\u00b0<\/p>\n<p>Question 18.<br>If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O. Such that \u2220C + \u2220D = k(\u2220AOB), then find the value of k.<br>Solution:<br>In quadrilateral ABCD,<br>Bisectors of \u2220A and \u2220B meet at O<br>Such that \u2220C + \u2220D = k (\u2220AOB)<br><img src=\"https:\/\/farm2.staticflickr.com\/1914\/30703632377_d9a33fbabb_o.png\" alt=\"Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions\" width=\"354\" height=\"726\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1916\/30703632217_e399cdb446_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables\" width=\"193\" height=\"55\"><\/p>\n<p>Question 19.<br>In the figure, PQRS is a rhombus in which the diagonal PR is produced to T. If \u2220SRT = 152\u00b0, find x, y and z.<br><img src=\"https:\/\/farm2.staticflickr.com\/1931\/30703632107_7bac856d32_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"216\" height=\"187\"><br>Solution:<br>In rhombus PQRS,<br>Diagonal PR and SQ bisect each other at right angles and PR is produced to T such that \u2220SRT = 152\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1902\/44730419825_734affb475_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"218\" height=\"191\"><br>But \u2220SRT + \u2220SRP = 180\u00b0 (Linear pair)<br>\u21d2 152\u00b0 +\u2220SRP = 180\u00b0<br>\u21d2 \u2220SRP =180\u00b0- 152\u00b0 = 28\u00b0<br>But \u2220SPR = \u2220SRP (\u2235 PR bisects \u2220P and \u2220R)<br>\u21d2 z = 28\u00b0<br>y = 90\u00b0 (\u2235 Diagonals bisect each other at right angles)<br>\u2220RPQ = z = 28\u00b0<br>\u2234 In \u2206POQ,<br>z + x = 90\u00b0 \u21d2 28\u00b0 + x = 90\u00b0<br>\u21d2 x = 90\u00b0 \u2013 28\u00b0 = 62\u00b0<br>\u2234 x = 62\u00b0, y = 90\u00b0, z = 28\u00b0<\/p>\n<p>Question 20.<br>In the figure, ABCD is a rectangle in which diagonal AC is produced to E. If \u2220ECD = 146\u00b0, find \u2220AOB.<br><img src=\"https:\/\/farm2.staticflickr.com\/1940\/44919528454_c31ef0f8ba_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"270\" height=\"165\"><br>Solution:<br>In rectangle ABCD,<br>Diagonals AC and BD bisect each other at O<br>AC is produced to E and \u2220DCE = 146\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1904\/44730419445_96d5d533c3_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 13 Linear Equations in Two Variables\" width=\"278\" height=\"166\"><br>\u2220DCE + \u2220DCA = 180\u00b0 (Linear pair)<br>\u21d2 146\u00b0+ \u2220DCA= 180\u00b0<br>\u21d2 \u2220DCA = 180\u00b0- 146\u00b0<br>\u21d2 \u2220DCA = 34\u00b0<br>\u2234 \u2220CAB = \u2220DCA (Alternate angles)<br>= 34\u00b0<br>Now in \u2206AOB,<br>\u2220AOB = 180\u00b0 \u2013 (\u2220DAB + \u2220OBA)<br>= 180\u00b0 \u2013 (34\u00b0 + 34\u00b0)<br>= 1803 \u2013 68\u00b0 = 112\u00b0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solution-chapter-13-linear-equations-in-two-variables-mcqs\"><\/span>RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables MCQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Mark the correct alternative in each of the following:<br>Question 1.<br>The opposite sides of a quadrilateral have<br>(a) no common point<br>(b) one common point<br>(c) two common points<br>(d) infinitely many common points<br>Solution:<br>The opposite sides of a quadrilateral have no common point. (a)<\/p>\n<p>Question 2.<br>The consecutive sides of a quadrilateral have<br>(a) no common point<br>(b) one common point<br>(c) two common points<br>(d) infinitely many common points<br>Solution:<br>The consecutive sides of a quadrilateral have one common point. (b)<\/p>\n<p>Question 3.<br>PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following cases, PQRS is a parallelogram?<br>(a) \u2220P = 100\u00b0, \u2220Q = 80\u00b0, \u2220R = 100\u00b0<br>(b) \u2220P = 85\u00b0, \u2220Q = 85\u00b0, \u2220R = 95\u00b0<br>(c) PQ = 7 cm, QR = 7 cm, RS = 8 cm, SP = 8 cm<br>(d) OP = 6.5 cm, OQ = 6.5 cm, OR = 5.2 cm, OS = 5.2 cm<br>Solution:<br>PQRS is a quadrilateral, PR and QS intersect each other at O. PQRS is a parallelogram if \u2220P = 100\u00b0, \u2220Q = 80\u00b0, \u2220R = 100\u00b0 (a)<\/p>\n<p>Question 4.<br>Which of the following quadrilateral is not a rhombus?<br>(a) All four sides are equal<br>(b) Diagonals bisect each other<br>(c) Diagonals bisect opposite angles<br>(d) One angle between the diagonals is 60\u00b0<br>Solution:<br>A quadrilateral is not a rhombus if one angle between the diagonals is 60\u00b0. (d)<\/p>\n<p>Question 5.<br>Diagonals necessarily bisect opposite angles in a<br>(a) rectangle<br>(b) parallelogram<br>(c) isosceles trapezium<br>(d) square<br>Solution:<br>Diagonals necessarily bisect opposite angles in a square. (d)<\/p>\n<p>Question 6.<br>The two diagonals are equal in a<br>(a) parallelogram<br>(b) rhombus<br>(c) rectangle<br>(d) trapezium<br>Solution:<br>The two diagonals are equal in a rectangle. (c)<\/p>\n<p>Question 7.<br>We get a rhombus by joining the mid-points of the sides of a<br>(a) parallelogram<br>(b) rhombus<br>(c) rectangle<br>(d) triangle<br>Solution:<br>We get a rhombus by joining the mid points of the sides of a rectangle. (c)<\/p>\n<p>Question 8.<br>The bisectors of any two adjacent angles of a parallelogram intersect at<br>(a) 30\u00b0<br>(b) 45\u00b0<br>(c) 60\u00b0<br>(d) 90\u00b0<br>Solution:<br>The bisectors of any two adjacent angles of a parallelogram intersect at 90\u00b0. (d)<\/p>\n<p>Question 9.<br>The bisectors of the angle of a parallelogram enclose a<br>(a) parallelogram<br>(b) rhombus<br>(c) rectangle<br>(d) square<br>Solution:<br>The bisectors of the angles of a parallelogram enclose a rectangle. (c)<\/p>\n<p>Question 10.<br>The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a<br>(a) parallelogram<br>(b) rectangle<br>(c) square<br>(d) rhombus<br>Solution:<br>The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. (a)<\/p>\n<p>Question 11.<br>The figure formed by joining the mid-points of the adjacent sides of a rectangle is a<br>(a) square<br>(b) rhombus<br>(c) trapezium<br>(d) none of these<br>Solution:<br>The figure formed by joining the mid-points of the adjacent sides of a rectangle is a rhombus. (b)<\/p>\n<p>Question 12.<br>The figure formed by joining the mid-points of the adjacent sides of a rhombus is a<br>(a) square<br>(b) rectangle<br>(c) trapezium<br>(d) none of these<br>Solution:<br>The figure formed by the joining the mid-points of the adjacent sides of a rhombus is a rectangle. (b)<\/p>\n<p>Question 13.<br>The figure formed by joining the mid-points of the adjacent sides of a square is a<br>(a) rhombus<br>(b) square<br>(c) rectangle<br>(d) parallelogram<br>Solution:<br>Tire figure formed by joining the mid-points of the adjacent sides of a square is a square. (b)<\/p>\n<p>Question 14.<br>The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a<br>(a) rectangle<br>(b) parallelogram<br>(b) rhombus<br>(d) square<br>Solution:<br>The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a parallelogram. (b)<\/p>\n<p>Question 15.<br>If one angle of a parallelogram is 24\u00b0 less than twice the smallest angle, then the measure of the largest angle of the parallelogram is<br>(a) 176\u00b0<br>(b) 68\u00b0<br>(c) 112\u00b0<br>(d) 102\u00b0<br>Solution:<br>Let the smallest angle be x<br>The largest angle = 2x \u2013 24\u00b0<br>But sum of two adjacent angles = 180\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1951\/45644579971_78f5e311cc_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables\" width=\"346\" height=\"126\"><\/p>\n<p>Question 16.<br>In a parallelogram ABCD, If \u2220DAB = 75\u00b0 and \u2220DBC = 60\u00b0, then \u2220BDC =<br>(a) 75\u00b0<br>(b) 60\u00b0<br>(c) 45\u00b0<br>(d) 55\u00b0<br>Solution:<br>In ||gm ABC,<br><img src=\"https:\/\/farm2.staticflickr.com\/1970\/45594030012_738f02465e_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"184\" height=\"168\"><br>\u2220A = 75\u00b0, \u2220DBC = 60\u00b0<br>But \u2220A + \u2220B = 180\u00b0 (Sum of two consecutive angles)<br>\u21d2 75\u00b0 + \u2220B = 180\u00b0<br>\u21d2 \u2220B = 180\u00b0- 75\u201c= 105\u00b0<br>But \u2220DBC = 60\u00b0<br>\u2234 \u2220DBA = 105\u00b0-60\u00b0 = 45\u00b0<br>But \u2220BDC = \u2220DBA (Alternate angles)<br>\u2234 \u2220BDC = 45\u00b0 (c)<\/p>\n<p>Question 17.<br>ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =<br>(a) AE<br>(b) BE<br>(c) CE<br>(d) DE<br>Solution:<br>In ||gm ABCD, BD is joined forming two triangles ABD and BCD<br><img src=\"https:\/\/farm2.staticflickr.com\/1903\/45644579831_3c4ec2ba38_o.png\" alt=\"Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions\" width=\"208\" height=\"157\"><br>E and F are the centroid of \u2206ABD and \u2206BCD<br>Now E and F trisect AC<br>i.e. AE = EF = FC<br>\u2234 EF = AE (a)<\/p>\n<p>Question 18.<br>ABCD is a parallelogram, M is the mid\u00acpoint of BD and BM bisects \u2220B. Then, \u2220AMB =<br>(a) 45\u00b0<br>(b) 60\u00b0<br>(c) 90\u00b0<br>(d) 75\u00b0<br>Solution:<br>In ||gm ABCD, M is mid-point of BD and<br>BM bisects \u2220B<br>AM is joined<br><img src=\"https:\/\/farm2.staticflickr.com\/1910\/45644579741_ae51aa8411_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables\" width=\"202\" height=\"160\"><br>\u2234AM bisects \u2220A<br>But \u2220A + \u2220B = 180\u00b0 (Sum of two consecutive angles)<br>\u2234 \u2220AMB = 90\u00b0 (c)<\/p>\n<p>Question 19.<br>If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is<br>(a) 108\u00b0<br>(b) 54\u00b0<br>(c) 12\u00b0<br>(d) 81\u00b0<br>Solution:<br>Let adjacent angle of a ||gm = x<br>Then second angle =&nbsp;<span id=\"MathJax-Element-47-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-297\" class=\"math\"><span id=\"MathJax-Span-298\" class=\"mrow\"><span id=\"MathJax-Span-299\" class=\"mfrac\"><span id=\"MathJax-Span-300\" class=\"mn\">2<\/span><span id=\"MathJax-Span-301\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;x<br>\u2234 x+&nbsp;<span id=\"MathJax-Element-48-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-302\" class=\"math\"><span id=\"MathJax-Span-303\" class=\"mrow\"><span id=\"MathJax-Span-304\" class=\"mfrac\"><span id=\"MathJax-Span-305\" class=\"mn\">2<\/span><span id=\"MathJax-Span-306\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;x= 180\u00b0<br>(Sum of two adjacent angles of a ||gm is 180\u00b0)<br><img src=\"https:\/\/farm2.staticflickr.com\/1915\/44730430685_a761311e08_o.png\" alt=\"RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables\" width=\"355\" height=\"312\"><\/p>\n<p>Question 20.<br>If the degree measures of the angles of quadrilateral are Ax, lx, 9x and 10JC, what is the sum of the measures of the smallest angle and largest angle?<br>(a) 140\u00b0<br>(b) 150\u00b0<br>(c) 168\u00b0<br>(d) 180\u00b0<br>Solution:<br>Sum of the angles of a quadrilateral = 360\u00b0<br>\u2234 4x + 1x + 9x + 10x = 360\u00b0<br>\u21d2 30x = 360\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-49-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-307\" class=\"math\"><span id=\"MathJax-Span-308\" class=\"mrow\"><span id=\"MathJax-Span-309\" class=\"mfrac\"><span id=\"MathJax-Span-310\" class=\"msubsup\"><span id=\"MathJax-Span-311\" class=\"texatom\"><span id=\"MathJax-Span-312\" class=\"mrow\"><span id=\"MathJax-Span-313\" class=\"mn\">360<\/span><\/span><\/span><span id=\"MathJax-Span-314\" class=\"texatom\"><span id=\"MathJax-Span-315\" class=\"mrow\"><span id=\"MathJax-Span-316\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-317\" class=\"mn\">30<\/span><\/span><\/span><\/span><\/span>&nbsp; = 12\u00b0<br>Now sum of smallest and largest angle = 4 x 12\u00b0 + 10 x 12\u00b0<br>= 48\u00b0 + 120\u00b0 = 168\u00b0 (c)<\/p>\n<p>Question 21.<br>If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to<br>(a) 16 cm<br>(b) 15 cm<br>(c) 20 cm<br>(d) 17 cm<br>Solution:<br>Diagonals of a rhombus are 18 cm and 24 cm But diagonals of a rhombus bisect each other at right angles<br><img src=\"https:\/\/farm2.staticflickr.com\/1928\/45644579651_75fa6793f5_o.png\" alt=\"Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions\" width=\"342\" height=\"430\"><\/p>\n<p>Question 22.<br>ABCD is a parallelogram in which diagonal AC bisects \u2220BAD. If \u2220BAC = 35\u00b0, then \u2220ABC =<br>(a) 70\u00b0<br>(b) 110\u00b0<br>(c) 90\u00b0<br>(d) 120\u00b0<br>Solution:<br>In ||gm ABCD, AC is its diagonal which bisect \u2220BAD<br><img src=\"https:\/\/farm2.staticflickr.com\/1963\/45644579421_0a53094b36_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables\" width=\"203\" height=\"150\"><br>\u2220BAD = 35\u00b0<br>\u2234 \u2220BAD = 2 x 35\u00b0 = 70\u00b0<br>But \u2220A + \u2220B = 180\u00b0 (Sum of consecutive angles)<br>\u21d2 70\u00b0 + \u2220B = 180\u00b0\u21d2 \u2220B = 180\u00b0 \u2013 70\u00b0<br>\u2234 \u2220B = 110\u00b0<br>\u21d2 ABC = 110\u00b0 (b)<\/p>\n<p>Question 23.<br>In a rhombus ABCD, if \u2220ACB = 40\u00b0, then \u2220ADB =<br>(a) 70\u00b0<br>(b) 45\u00b0<br>(c) 50\u00b0<br>(d) 60\u00b0<br>Solution:<br>In rhombus ABCD, \u2220ACB = 40\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1930\/44730430185_c27d11917f_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"167\" height=\"151\"><br>\u2234 \u2220BCD = 2 x \u2220ACB<br>= 2 x 40\u00b0 = 80\u00b0<br>But \u2220BCD + \u2220ADC = 180\u00b0 (Sum of consecutive angles of ||gm)<br>\u21d2 80\u00b0 + \u2220ADC = 180\u00b0<br>\u21d2 \u2220ADC = 180\u00b0 \u2013 80\u00b0 = 100\u00b0<br>\u2234 \u2220ADB =&nbsp;<span id=\"MathJax-Element-50-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-318\" class=\"math\"><span id=\"MathJax-Span-319\" class=\"mrow\"><span id=\"MathJax-Span-320\" class=\"mfrac\"><span id=\"MathJax-Span-321\" class=\"mn\">1<\/span><span id=\"MathJax-Span-322\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220ADC =&nbsp;<span id=\"MathJax-Element-51-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-323\" class=\"math\"><span id=\"MathJax-Span-324\" class=\"mrow\"><span id=\"MathJax-Span-325\" class=\"mfrac\"><span id=\"MathJax-Span-326\" class=\"mn\">1<\/span><span id=\"MathJax-Span-327\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>x 100\u00b0 = 50\u00b0 (c)<\/p>\n<p>Question 24.<br>In \u2206ABC, \u2220A = 30\u00b0, \u2220B = 40\u00b0 and \u2220C = 110\u00b0. The angles of the triangle formed by joining the mid-points of the sides of this triangle are<br>(a) 70\u00b0, 70\u00b0, 40\u00b0<br>(b) 60\u00b0, 40\u00b0, 80\u00b0<br>(c) 30\u00b0, 40\u00b0, 110\u00b0<br>(d) 60\u00b0, 70\u00b0, 50\u00b0<br>Solution:<br>In \u2206ABC,<br><img src=\"https:\/\/farm2.staticflickr.com\/1912\/45644579341_d5a155aedb_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"266\" height=\"209\"><br>\u2220A = 30\u00b0, \u2220B = 40\u00b0, \u2220C = 110\u00b0<br>D, E and F are mid-points of the sides of the triangle. By joining them in order,<br>DEF is a triangle formed<br>Now BDEF, CDFE and AFDE are ||gms<br>\u2234 \u2220A = \u2220D = 30\u00b0<br>\u2220B = \u2220E = 40\u00b0<br>\u2220C = \u2220F= 110\u00b0<br>\u2234 Angles are 30\u00b0, 40\u00b0, 110\u00b0 (c)<\/p>\n<p>Question 25.<br>The diagonals of a parallelogram ABCD intersect at O. If \u2220BOC = 90\u00b0 and \u2220BDC = 50\u00b0, then \u2220OAB =<br>(a) 40\u00b0<br>(b) 50\u00b0<br>(c) 10\u00b0<br>(d) 90\u00b0<br>Solution:<br>In ||gm ABCD, diagonals AC and BD intersect each other at O<br><img src=\"https:\/\/farm2.staticflickr.com\/1969\/44730429895_6e47b5f23d_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"188\" height=\"179\"><br>BOC = 90\u00b0, \u2220BDC = 50\u00b0<br>\u2235 \u2220BOC = 90\u00b0<br>\u2234 Diagonals of ||gm bisect each other at 90\u00b0<br>\u2234\u2220COD = 90\u00b0<br>In \u2206COD,<br>\u2220OCD = 90\u00b0 \u2013 50\u00b0 = 40\u00b0<br>But \u2220OAB = \u2220OCD (Alternate angles)<br>\u2234\u2220OAB = 40\u00b0 (a)<\/p>\n<p>Question 26.<br>ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and BC respectively. If AB = 12 cm, MN = 14 cm, then CD =<br>(a) 10 cm<br>(b) 12 cm<br>(c) 14 cm<br>(d) 16 cm<br>Solution:<br>In trapezium AB || DC<br>M and N are mid-points of sides AD and BC and MN are joined<br>AB = 12 cm, MN = 14 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1954\/45644579251_d5c6417aba_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 13 Linear Equations in Two Variables\" width=\"211\" height=\"173\"><br>\u2235 MN =&nbsp;<span id=\"MathJax-Element-52-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-328\" class=\"math\"><span id=\"MathJax-Span-329\" class=\"mrow\"><span id=\"MathJax-Span-330\" class=\"mfrac\"><span id=\"MathJax-Span-331\" class=\"mn\">1<\/span><span id=\"MathJax-Span-332\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>(AB + CD)<br>\u21d2 2MN = AB + CD<br>\u21d2 2 x 14 = 12 + CD<br>CD = 2 x 14 \u2013 12 = 28 \u2013 12 = 16 cm (d)<\/p>\n<p>Question 27.<br>Diagonals of a quadrilateral ABCD bisect each other. If \u2220A = 45\u00b0, then \u2220B =<br>(a) 115\u00b0<br>(b) 120\u00b0<br>(c) 125\u00b0<br>(d) 135\u00b0<br>Solution:<br>Diagonals AC and BD of quadrilateral ABCD bisect each other at O<br><img src=\"https:\/\/farm2.staticflickr.com\/1901\/44730429565_d267258a2f_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"195\" height=\"168\"><br>\u2234 AO = OC, BO = OD<br>\u2234 ABCD is a ||gm \u2220A = 45\u00b0<br>But \u2220A + \u2220B = 180\u00b0 (Sum of consecutive angles)<br>\u2234 \u2220B = 180\u00b0 \u2013 \u2220A = 180\u00b0 \u2013 45\u00b0<br>= 135\u00b0 (d)<\/p>\n<p>Question 28.<br>P is the mid-point of side BC of a paralleogram ABCD such that \u2220BAP = \u2220DAP. If AD = 10 cm, then CD =<br>(a) 5 cm<br>(b) 6 cm<br>(c) 8 cm<br>(d) 10 cm<br>Solution:<br>In ||gm ABCD, P is mid-point of BC<br>AD = 10cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1951\/45644579181_8e01ece027_o.png\" alt=\"RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables \" width=\"207\" height=\"242\"><br>\u2220BAP = \u2220DAP<br>Through P, draw PQ || AB<br>\u2234 ABPQ is rhombus<br>\u2234 AB = BP = AQ<br>=&nbsp;<span id=\"MathJax-Element-53-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-333\" class=\"math\"><span id=\"MathJax-Span-334\" class=\"mrow\"><span id=\"MathJax-Span-335\" class=\"mfrac\"><span id=\"MathJax-Span-336\" class=\"mn\">1<\/span><span id=\"MathJax-Span-337\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AB =&nbsp;<span id=\"MathJax-Element-54-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-338\" class=\"math\"><span id=\"MathJax-Span-339\" class=\"mrow\"><span id=\"MathJax-Span-340\" class=\"mfrac\"><span id=\"MathJax-Span-341\" class=\"mn\">1<\/span><span id=\"MathJax-Span-342\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;x 10 = 5 cm<br>But CD = AB (Opposite sides of ||gm)<br>\u2234 CD = 5 cm (a)<\/p>\n<p>Question 29.<br>In \u2206ABC, E is the mid-point of median AD such that BE produced meets AC at E If AC = 10.5 cm, then AF =<br>(a) 3 cm<br>(b) 3.5 cm<br>(c) 2.5 cm<br>(d) 5 cm<br>Solution:<br>In \u2206ABC, E is the mid-point of median AD<br>Such that BE produced meets AC at F<br>AC = 10.5 cm<br>Draw DG || AF<br><img src=\"https:\/\/farm2.staticflickr.com\/1907\/44730429365_89a06ded26_o.png\" alt=\"RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables MCQS\" width=\"220\" height=\"188\"><br>In \u2206ADG<br>E is mid-point of AD and EF || DG<br>\u2234 F is mid-point of AG<br>\u21d2 AF = FG \u2026(i)<br>In \u2206BCF<br>D is mid-point of BC and DG || BF<br>\u2234 G is mid-point of FC<br>\u2234 FG = GC \u2026(i)<br>From (i) and (ii)<br>AF = FG = GC =&nbsp;<span id=\"MathJax-Element-55-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-343\" class=\"math\"><span id=\"MathJax-Span-344\" class=\"mrow\"><span id=\"MathJax-Span-345\" class=\"mfrac\"><span id=\"MathJax-Span-346\" class=\"mn\">1<\/span><span id=\"MathJax-Span-347\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;AC<br>But AC = 10.5 cm<br>\u2234 AF =&nbsp;<span id=\"MathJax-Element-56-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-348\" class=\"math\"><span id=\"MathJax-Span-349\" class=\"mrow\"><span id=\"MathJax-Span-350\" class=\"mfrac\"><span id=\"MathJax-Span-351\" class=\"mn\">1<\/span><span id=\"MathJax-Span-352\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;AC =&nbsp;<span id=\"MathJax-Element-57-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-353\" class=\"math\"><span id=\"MathJax-Span-354\" class=\"mrow\"><span id=\"MathJax-Span-355\" class=\"mfrac\"><span id=\"MathJax-Span-356\" class=\"mn\">1<\/span><span id=\"MathJax-Span-357\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;x 10.5 = 3.5 cm (b)<\/p>\n<p>Question 30.<br>ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =<br><img src=\"https:\/\/farm2.staticflickr.com\/1971\/45594027942_db2f955a1a_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"267\" height=\"106\"><br>Solution:<br>In ||gm ABCD, E is mid-point of BC DE and AB are produced to meet at F<br><img src=\"https:\/\/farm2.staticflickr.com\/1931\/45594027842_e5243944ff_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"309\" height=\"149\"><br>\u2235 E is mid point of BC<br>\u2234 BE = EC<br>In \u2206BEF and \u2206CDE<br>BE = EC<br>\u2220BEF = \u2220CED (Vertically opposite angle)<br>and \u2220EBF = \u2220ECD (Alternate angles)<br>\u2234 \u2206BEF \u2245 \u2206CDE (ASA criterian)<br>\u2234 DC = BF<br>But DC = AB<br>\u2234 AB = BF<br>AF = AB + BF = AB + AB<br>= 2AB (b)<\/p>\n<p>Question 31.<br>In a quadrilateral ABCD, \u2220A + \u2220C is 2 times \u2220B + \u2220D. If \u2220A = 140\u00b0 and \u2220D = 60\u00b0, then \u2220B =<br>(a) 60\u00b0<br>(b) 80\u00b0<br>(c) 120\u00b0<br>(d) None of these<br>Solution:<br>In quadrilateral ABCD<br>\u21d2 \u2220A + \u2220C = 2(\u2220B + \u2220D)<br>\u21d2 \u2220A + \u2220C = 2\u2220B + 2\u2220D<br>Adding 2\u2220A + 2\u2220C both sides<br>2\u2220A + 2\u2220C + \u2220A + \u2220C = 2\u2220A + 2\u2220B + 2\u2220C + 2\u2220D<br>\u21d2 3\u2220A + 3\u2220C = 2(\u2220A + \u2220B + \u2220C + \u2220D)<br>\u21d2 3(\u2220A + \u2220C) = 2 x 360\u00b0 = 720\u00b0<br>\u2234 \u2220A + \u2220C =&nbsp;<span id=\"MathJax-Element-58-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-358\" class=\"math\"><span id=\"MathJax-Span-359\" class=\"mrow\"><span id=\"MathJax-Span-360\" class=\"mfrac\"><span id=\"MathJax-Span-361\" class=\"msubsup\"><span id=\"MathJax-Span-362\" class=\"texatom\"><span id=\"MathJax-Span-363\" class=\"mrow\"><span id=\"MathJax-Span-364\" class=\"mn\">720<\/span><\/span><\/span><span id=\"MathJax-Span-365\" class=\"texatom\"><span id=\"MathJax-Span-366\" class=\"mrow\"><span id=\"MathJax-Span-367\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-368\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp; = 240\u00b0<br>\u21d2 40\u00b0 + \u2220C = 240\u00b0 (\u2235 \u2220A = 40\u00b0)<br>\u2220C = 240\u00b0 \u2013 40\u00b0 = 200\u00b0<br>Now 2(\u2220B + \u2220D) = \u2220A + \u2220C = 240\u00b0<br>\u2220B + \u2220D =&nbsp;<span id=\"MathJax-Element-59-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-369\" class=\"math\"><span id=\"MathJax-Span-370\" class=\"mrow\"><span id=\"MathJax-Span-371\" class=\"mfrac\"><span id=\"MathJax-Span-372\" class=\"msubsup\"><span id=\"MathJax-Span-373\" class=\"texatom\"><span id=\"MathJax-Span-374\" class=\"mrow\"><span id=\"MathJax-Span-375\" class=\"mn\">240<\/span><\/span><\/span><span id=\"MathJax-Span-376\" class=\"texatom\"><span id=\"MathJax-Span-377\" class=\"mrow\"><span id=\"MathJax-Span-378\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-379\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 120\u00b0<br>\u2234 \u2220B = 60\u00b0 = 120\u00b0<br>\u2234 \u2220B = 60\u00b0 (a)<\/p>\n<p>Question 32.<br>The diagonals AC and BD of a rectangle ABCD intersect each other at P. If \u2220ABD = 50\u00b0, then \u2220DPC =<br>(a) 70\u00b0<br>(b) 90\u00b0<br>(c) 80\u00b0<br>(d) 100\u00b0<br>Solution:<br>In rectangle ABCD, diagonals AC and BD intersect each other at P<br><img src=\"https:\/\/farm2.staticflickr.com\/1920\/44730428765_2069b5d697_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables\" width=\"179\" height=\"165\"><br>\u2220ABD = 50\u00b0<br>\u2234 \u2220CAB = \u2220ABD = 50\u00b0 (\u2235 AP = BP)<br>Now in \u2206APB<br>\u2220CAB + \u2220ABD + \u2220APB = 180\u00b0 (Angles of a triangle)<br>\u21d2 \u2220PAB + \u2220PBA + \u2220APB = 180\u00b0<br>\u21d2 50\u00b0 + 50\u00b0 + \u2220APB = 180\u00b0<br>\u21d2 \u2220APB = 180\u00b0 \u2013 50\u00b0 \u2013 50\u00b0 = 80\u00b0<br>But \u2220DPC = ADB (Vertically opposite angles)<br>\u2234 \u2220DPC = 80\u00b0 (c)<\/p>\n<p>RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables Ex 13.1 Q1<\/p>\n<p><img class=\"alignnone size-full wp-image-6642\" src=\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2014\/12\/RD-Sharma-Class-9-Solutions-Chapter-13-Linear-Equations-in-Two-Variables-1.png\" sizes=\"(max-width: 471px) 100vw, 471px\" srcset=\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2014\/12\/RD-Sharma-Class-9-Solutions-Chapter-13-Linear-Equations-in-Two-Variables-1.png 471w, https:\/\/www.learncbse.in\/wp-content\/uploads\/2014\/12\/RD-Sharma-Class-9-Solutions-Chapter-13-Linear-Equations-in-Two-Variables-1-211x300.png 211w\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 1\" width=\"471\" height=\"668\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8615\/16077393375_92cef4f212_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 2.\" width=\"447\" height=\"682\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8602\/15891306269_7a6cbb2135_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 3\" width=\"395\" height=\"589\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7506\/16051595616_bfe8bec0b0_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 4.\" width=\"431\" height=\"288\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8562\/15890106890_5c86394f54_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 5.\" width=\"462\" height=\"605\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8568\/16076692792_881d4489a4_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 6.\" width=\"451\" height=\"436\"><br><img class=\"alignnone size-full wp-image-6641\" src=\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2014\/12\/RD-Sharma-Class-9-Solutions-Chapter-13-Linear-Equations-in-Two-Variables-7.png\" sizes=\"(max-width: 386px) 100vw, 386px\" srcset=\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2014\/12\/RD-Sharma-Class-9-Solutions-Chapter-13-Linear-Equations-in-Two-Variables-7.png 386w, https:\/\/www.learncbse.in\/wp-content\/uploads\/2014\/12\/RD-Sharma-Class-9-Solutions-Chapter-13-Linear-Equations-in-Two-Variables-7-300x289.png 300w\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 7\" width=\"386\" height=\"372\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7555\/15891305319_1ec94ce79f_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 8.\" width=\"465\" height=\"583\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7480\/15891629317_f352933cc4_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 9.\" width=\"458\" height=\"331\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7553\/15891628937_f192bd4b2f_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 10.\" width=\"519\" height=\"686\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7575\/15889950978_caef2263e6_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 11\" width=\"450\" height=\"306\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8588\/16077390895_923a929433_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 12\" width=\"518\" height=\"679\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7524\/16077390295_bdccbe2474_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 13\" width=\"471\" height=\"274\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7473\/15455087344_54bb93cbd2_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 14.\" width=\"519\" height=\"652\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8584\/15889949718_241539f18e_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 15.\" width=\"456\" height=\"279\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7577\/15890103750_4af78c1240_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 16.\" width=\"526\" height=\"684\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7512\/15890103470_c621fd13a4_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 17\" width=\"446\" height=\"418\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8587\/16076689282_a5ac127301_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 18\" width=\"520\" height=\"696\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8608\/15889948438_7fac7d62fd_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 19\" width=\"459\" height=\"565\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7475\/16051590706_a069cb1385_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 20\" width=\"526\" height=\"690\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8569\/15890101870_0252edc851_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 21\" width=\"475\" height=\"547\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7551\/16051589946_d570d60d51_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 22\" width=\"515\" height=\"701\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7569\/15455084064_e794e4cd94_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 23\" width=\"476\" height=\"334\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8568\/16077386665_11bf968eb7_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 24\" width=\"523\" height=\"699\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7495\/15457725803_9e6570824d_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 25\" width=\"486\" height=\"582\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7528\/15891298849_0280d064d2_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 26\" width=\"459\" height=\"682\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8606\/15457724973_02aab79659_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 27\" width=\"461\" height=\"656\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8674\/15889944678_b7744faa3c_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 28\" width=\"456\" height=\"673\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7561\/16051586926_9e92237b43_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 29\" width=\"464\" height=\"591\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7569\/16051613826_630ba8ac1c_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 30\" width=\"453\" height=\"700\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8600\/16077411275_393df4b8cd_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 31\" width=\"471\" height=\"693\"><br>RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables<br><img src=\"https:\/\/farm8.staticflickr.com\/7475\/15891648857_390a803212_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 32\" width=\"505\" height=\"695\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8661\/15890124040_75c4a6ac28_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 33\" width=\"522\" height=\"694\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7555\/15455107454_4235f5bb9a_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 34\" width=\"511\" height=\"669\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7498\/15455107014_076220603d_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 35\" width=\"463\" height=\"517\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7466\/15890123220_6b50777392_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 36\" width=\"525\" height=\"698\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8627\/15890122570_fbe367f6f4_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 37\" width=\"499\" height=\"590\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7467\/15457748933_74fd4c2307_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 38\" width=\"488\" height=\"468\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8624\/16077407945_6ce05d9eb5_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 39\" width=\"512\" height=\"684\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7489\/15891645597_000f856042_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 40\" width=\"465\" height=\"362\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7498\/15891645267_a5827eac62_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 41\" width=\"518\" height=\"679\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7510\/15890120300_929402e5e5_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 42\" width=\"458\" height=\"333\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7580\/15891319699_c279c06a1c_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 43\" width=\"526\" height=\"693\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7483\/15890119440_61b248eb44_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 44\" width=\"464\" height=\"399\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7548\/15455102834_00c9273e4b_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 45\" width=\"519\" height=\"688\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7551\/15890118530_0829e19815_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 46\" width=\"458\" height=\"384\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7516\/15889965008_ab8bf381d5_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 47\" width=\"529\" height=\"691\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7517\/15455101544_495f645365_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 48\" width=\"485\" height=\"527\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7534\/15457744113_f3e989ec1b_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 49\" width=\"522\" height=\"729\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8616\/16077402725_6345844fdf_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 50\" width=\"469\" height=\"373\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7491\/16076703692_9c6a83d267_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 51\" width=\"515\" height=\"684\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7579\/16076703342_585c2106d9_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 52\" width=\"487\" height=\"317\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8623\/15891315409_21ca5e4861_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 53\" width=\"521\" height=\"687\"><br>CBSE RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables Exercise problems<br><img src=\"https:\/\/farm9.staticflickr.com\/8683\/16077401515_120d6f768b_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 54\" width=\"507\" height=\"694\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8676\/15455098564_df8998fac4_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 55\" width=\"526\" height=\"706\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7522\/15455098174_0c262ff27e_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 56\" width=\"468\" height=\"430\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7559\/15891638357_9e755c0dd1_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 57\" width=\"470\" height=\"685\"><br>CBSE RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables Exercise problems<br><img src=\"https:\/\/farm9.staticflickr.com\/8676\/15457740593_6646cf96fa_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 58\" width=\"511\" height=\"703\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7582\/15891637617_8ac4be8d4b_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 59\" width=\"485\" height=\"397\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7505\/15457740133_3813c4c626_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 60\" width=\"485\" height=\"703\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7580\/16051602356_82f1670bf5_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 61\" width=\"521\" height=\"689\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7561\/15455096174_a892568e54_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 62\" width=\"500\" height=\"702\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7519\/15455095634_e2da534f72_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 63\" width=\"521\" height=\"716\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7496\/15891311149_e7a5a8c2af_o.png\" alt=\"solutions for Class 9 Chapter 13 Linear Equations in Two Variables 64\" width=\"465\" height=\"455\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7503\/15891311039_0342389c96_o.png\" alt=\"RD Sharma solutions for Class 9 Chapter 13 Linear Equations in Two Variables 65\" width=\"512\" height=\"685\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7510\/15457737653_4ef0e3028b_o.png\" alt=\"RD Sharma solutions for Class 9 Chapter 13 Linear Equations in Two Variables 66\" width=\"459\" height=\"597\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7532\/16051600016_f0955723d0_o.png\" alt=\"RD Sharma solutions for Class 9 Chapter 13 Linear Equations in Two Variables 67\" width=\"465\" height=\"621\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8574\/15455094174_6763ff5841_o.png\" alt=\"RD Sharma solutions for Class 9 Chapter 13 Linear Equations in Two Variables 68\" width=\"445\" height=\"621\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7526\/15891309829_19f4273e4d_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 69\" width=\"453\" height=\"636\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7559\/15457736443_bd76e591fe_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 70\" width=\"446\" height=\"589\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7538\/16051598706_bb7e9b7e2a_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 71\" width=\"428\" height=\"310\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7553\/16077395595_74fd3fd4db_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 72\" width=\"518\" height=\"683\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7564\/15890109290_c7edd189da_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 73\" width=\"456\" height=\"405\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7557\/15455092094_5a67aa826b_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 74\" width=\"517\" height=\"677\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8628\/16077394415_33b4ec2c79_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables 75\" width=\"478\" height=\"508\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7525\/15890108500_a1d1c26ac3_o.png\" alt=\"RD Sharma solutions for Class 9 Chapter 13 Linear Equations in Two Variables 76\" width=\"458\" height=\"495\"><\/p>\n<h2 style=\"text-align: left;\"><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-important-topics\"><\/span><strong>Detailed Exercise-wise Explanation with Important Topics<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-13a\"><\/span>RD Sharma Solutions Class 9 Chapter 13A<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the Class 9 RD Sharma Solutions Chapter 10 Exercise 10C the students will find step by step explanations which have been done by different experts. The main areas which the experts have tried to focus on are problems and a brief introduction of linear equations in two variables. Once you examine the contents you will be able to figure out the parts which need your immediate attention.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-13b\"><\/span>RD Sharma Solutions Class 9 Chapter 13B<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the RD Sharma Class 9 Chapter 10 Exercise 10D Solutions you will be exposed to all kinds of tricky problems from Linear Equation in Two Variables. Try to answer all the questions from RD Sharma Class 9 Chapter 13B without seeing the answers from Chapter 13 solutions. Access your performance and make a note of it. After some time you check all the answers from RD Sharma Solutions Class 9 Chapter 13B.<\/p>\n<p>If you make a habit you will get the maximum benefit in your final Maths exam. The main focus of RD Sharma Class 9 Solutions Chapter 13 Exercise 13B is to solve problems based on a linear equation. You will be in a commendable position if you look for the little things at the time of practicing questions from Class 9 RD Sharma Chapter 13B.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-13c\"><\/span>RD Sharma Solutions Class 9 Chapter 13C<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the RD Sharma Solutions Class 9 Maths Chapter 13, the students will get all kind of explanations which will clear the basic concepts of Linear Equation in Two Variables chapter. The main area which has been dealt with here are connected using graphical representations.<\/p>\n<p>The step-wise explanations have been provided by the experts to satisfy the needs of the students. The important concepts will become crystal clear and the students will be confident enough to solve their own problems. The students are required to solve problems<br>on a daily basis for maximum benefit.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-13d\"><\/span>RD Sharma Solutions Class 9 Chapter 13D<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the exercise 13D of RD Sharma Solutions, students will get to know about the techniques which can be applied in case of Linear Equation in Two Variables. Parts like Equations of lines parallel to the x-axis and y-axis have been addressed by the experts with an extra bit of care and all you need to do are sit with the relevant parts after understanding the concepts.<\/p>\n<p>You can also include the parts in your study materials while examining the essential aspects of Linear Equation in Two Variables. In the last exercise, you will have to focus on the various kinds of questions from Linear Equation in Two Variables. Often students face problems while answering objective-based questions and the RD Sharma Class 9 solutions will bring you back on track.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"important-concepts-rd-sharma-class-9-maths\"><\/span>Important concepts RD Sharma Class 9 Maths<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul>\n<li>Linear Equations in Two Variables Introduction.<\/li>\n<li>The solution of a linear equation.<\/li>\n<li>Graph of a linear equation in two variables.<\/li>\n<li>Equations of lines parallel to the x-axis and y-axis.<\/li>\n<\/ul>\n<p>This is the complete blog of <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 13. To know more about the CBSE Class 9 Maths, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-9-maths-chapter-13\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 13<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630751218988\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-13\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 13?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630751250503\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-13\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 13?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630751275853\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-13-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 13\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 9 Maths Chapter 13 &#8211; Linear Equations In Two Variables: Are you wondering about the methods which will be effective for your performance in Class 9 Annual Exam? You don\u2019t have to worry much about that because we are going to present you RD Sharma Solutions Class 9 Maths&nbsp;Chapter 13. Download &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 13 &#8211; Linear Equations In Two Variables (Updated for 2023-24)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 13 &#8211; Linear Equations In Two Variables (Updated for 2023-24)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124489,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3081,3037,3086],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62877"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=62877"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62877\/revisions"}],"predecessor-version":[{"id":473851,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62877\/revisions\/473851"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124489"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=62877"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=62877"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=62877"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}